Recently Adaltik discovered japanese crosswords. Japanese crossword is a picture, represented as a table sized a × b squares, and each square is colored white or black. There are integers to the left of the rows and to the top of the columns, encrypting the corresponding row or column. The number of integers represents how many groups of black squares there are in corresponding row or column, and the integers themselves represents the number of consecutive black squares in corresponding group (you can find more detailed explanation in Wikipedia https://en.wikipedia.org/wiki/Japanese_crossword).

Adaltik decided that the general case of japanese crossword is too complicated and drew a row consisting of n squares (e.g. japanese crossword sized 1 × n), which he wants to encrypt in the same way as in japanese crossword.

The example of encrypting of a single row of japanese crossword.

Help Adaltik find the numbers encrypting the row he drew.

The first line should contain a single integer k — the number of integers encrypting the row, e.g. the number of groups of black squares in the row.

The second line should contain k integers, encrypting the row, e.g. corresponding to sizes of groups of consecutive black squares in the order from left to right.

The last sample case correspond to the picture in the statement.

// <A.cpp> - Fri Sep 30 22:03:52 2016
// This file is made by YJinpeng，created by XuYike's black technology automatically.
// Copyright (C) 2016 ChangJun High School, Inc.
// I don't know what this program is.

#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <cmath>
using namespace std;
typedef long long LL;
const int MAXN=110;
inline LL gi() {register LL w=0,q=0;register char ch=getchar();while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();if(ch=='-')q=1,ch=getchar();while(ch>='0'&&ch<='9')w=w*10+ch-'0',ch=getchar();return q?-w:w;
}
char s[MAXN];int a[MAXN];
int main()
{freopen("A.in","r",stdin);freopen("A.out","w",stdout);int n=gi(),k=0;scanf("%s",s);for(int i=0;i<n;i++)if(s[i]=='B'){int j=i;k++;while(s[j++]=='B')a[k]++;i=j-1;}printf("%d\n",k);for(int i=1;i<=k;i++)printf("%d ",a[i]);return 0;
}

Print two integers — time (in seconds), Vanya needs to be authorized to Codehorses in the best case for him and in the worst case respectively.

Consider the first sample case. As soon as all passwords have the same length, Vanya can enter the right password at the first try as well as at the last try. If he enters it at the first try, he spends exactly 1 second. Thus in the best case the answer is 1. If, at the other hand, he enters it at the last try, he enters another 4 passwords before. He spends 2 seconds to enter first 2 passwords, then he waits 5 seconds as soon as he made 2 wrong tries. Then he spends 2 more seconds to enter 2 wrong passwords, again waits 5 seconds and, finally, enters the correct password spending 1 more second. In summary in the worst case he is able to be authorized in 15 seconds.

Consider the second sample case. There is no way of entering passwords and get the access to the site blocked. As soon as the required password has length of 2, Vanya enters all passwords of length 1 anyway, spending 2 seconds for that. Then, in the best case, he immediately enters the correct password and the answer for the best case is 3, but in the worst case he enters wrong password of length 2 and only then the right one, spending 4 seconds at all.

// <B.cpp> - Fri Sep 30 22:03:53 2016
// This file is made by YJinpeng，created by XuYike's black technology automatically.
// Copyright (C) 2016 ChangJun High School, Inc.
// I don't know what this program is.

#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#define MOD 1000000007
#define INF 1e9
#define IN inline
#define RG register
using namespace std;
typedef long long LL;
typedef long double LB;
const int MAXN=110;
const int MAXM=100010;
inline int max(int &x,int &y) {return x>y?x:y;}
inline int min(int &x,int &y) {return x<y?x:y;}
inline LL gi() {register LL w=0,q=0;register char ch=getchar();while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();if(ch=='-')q=1,ch=getchar();while(ch>='0'&&ch<='9')w=w*10+ch-'0',ch=getchar();return q?-w:w;
}
char c[MAXN];
struct node{char s[MAXN];int l;bool operator<(node b)const{return l<b.l;}void read(){scanf("%s",s);l=strlen(s);}bool pan(){for(int i=0;i<l;i++)if(s[i]!=c[i])return true;return false;}
}a[MAXN];
int main()
{freopen("B.in","r",stdin);freopen("B.out","w",stdout);int n=gi(),k=gi();for(int i=1;i<=n;i++)a[i].read();scanf("%s",c);int i,l=strlen(c),ans=0;sort(a+1,a+1+n);for(i=1;i<=n;i++){if(a[i].l>=l)break;ans++;if(i%k==0)ans+=5;}printf("%d ",ans+1);int kk=(i-1)%k;for(;i<=n;i++){if(a[i].l>l)break;if(a[i].pan()){ans++;kk++;if(kk%k==0)ans+=5;}}printf("%d",ans+1);return 0;
}

Print the single integer k (2 ≤ k ≤ n) — the maximum number of showplaces that Irina can visit during her journey from showplace 1 to showplace n within time not exceeding T, in the first line.

Print k distinct integers in the second line — indices of showplaces that Irina will visit on her route, in the order of encountering them.

If there are multiple answers, print any of them.

// <C.cpp> - Fri Sep 30 22:03:53 2016
// This file is made by YJinpeng，created by XuYike's black technology automatically.
// Copyright (C) 2016 ChangJun High School, Inc.
// I don't know what this program is.

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#define IN inline
#define RG register
using namespace std;
typedef long long LL;
const int MAXN=5010;
const int MAXM=MAXN<<1;
inline LL gi() {register LL w=0,q=0;register char ch=getchar();while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();if(ch=='-')q=1,ch=getchar();while(ch>='0'&&ch<='9')w=w*10+ch-'0',ch=getchar();return q?-w:w;
}
int n,tot,be,T;
int to[MAXM],ne[MAXM],fr[MAXN],W[MAXM];
int f[MAXN][MAXN],d[MAXN][MAXN];bool used[MAXN];
void add(int u,int v,int w){to[++tot]=v;ne[tot]=fr[u];fr[u]=tot;W[tot]=w;
}
void dfs(int x){if(x==n){f[x][1]=0;return;}if(used[x])return;used[x]=true;for(int i=fr[x];i;i=ne[i]){dfs(to[i]);for(int j=1;j<=n;j++){if(f[to[i]][j-1]==be)continue;if(f[to[i]][j-1]+W[i]>T)continue;//..会爆intif(f[to[i]][j-1]+W[i]<f[x][j]){f[x][j]=f[to[i]][j-1]+W[i];d[x][j]=to[i];}}}
}
int main()
{freopen("C.in","r",stdin);freopen("C.out","w",stdout);n=gi();int m=gi();T=gi();for(int i=1;i<=m;i++){int u=gi(),v=gi(),w=gi();add(u,v,w);}memset(f,123,sizeof(f));be=f[1][1];dfs(1);int i;for(i=n;i;i--)if(f[1][i]<=T)break;printf("%d\n",i);for(int o=1;i;i--){printf("%d ",o);o=d[o][i];}return 0;
}

Print n integers b₁, b₂, ..., b_n in the only line — the array elements after applying no more than k operations to the array. In particular, should stay true for every 1 ≤ i ≤ n, but the product of all array elements should be minimum possible.

If there are multiple answers, print any of them.

// <D.cpp> - Fri Sep 30 22:03:53 2016
// This file is made by YJinpeng，created by XuYike's black technology automatically.
// Copyright (C) 2016 ChangJun High School, Inc.
// I don't know what this program is.

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <queue>
#define IN inline
#define RG register
using namespace std;
typedef long long LL;
const int MAXN=200010;
inline LL gi() {register LL w=0,q=0;register char ch=getchar();while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();if(ch=='-')q=1,ch=getchar();while(ch>='0'&&ch<='9')w=w*10+ch-'0',ch=getchar();return q?-w:w;
}
LL a[MAXN];
struct node{LL s;int p;};
priority_queue<node>q;
IN bool operator<(RG node a,RG node b){return a.s>b.s;}
int main()
{freopen("D.in","r",stdin);freopen("D.out","w",stdout);int n=gi(),k=gi(),x=gi(),b=0;for(int i=1;i<=n;i++)a[i]=gi(),q.push((node){a[i]<0?-a[i]:a[i],i}),b+=a[i]<0;while(k--){int p=q.top().p;q.pop();b-=a[p]<0;if(b&1)a[p]+=x;else a[p]-=x;b+=a[p]<0;q.push((node){a[p]<0?-a[p]:a[p],p});}for(int i=1;i<=n;i++)printf("%lld ",a[i]);return 0;
}