#include<bits/stdc++.h>
using namespace std;
const int MAXN=3007;
int n,m,i,j,k;
long long ans=1e18,cnt;
priority_queue<int,vector<int>,greater<int> >e[MAXN],s[MAXN],S;
int main()
{cin>>n>>m;for(i=1; i<=n; ++i){scanf("%d%d",&j,&k);s[j].push(k);}for(i=1; i<=n; ++i){while(!S.empty())S.pop();cnt=s[1].size();for(j=2; j<=m; ++j)e[j]=s[j];long long v=0;for(j=2; j<=m; ++j){while(e[j].size()>=i){v+=e[j].top();e[j].pop();cnt++;}}for(j=2; j<=m; ++j){while(!e[j].empty()){S.push(e[j].top());e[j].pop();}}for(; cnt<i&&!S.empty(); ++cnt){v+=S.top();S.pop();}if(cnt>=i) ans=min(ans,v);}cout<<ans<<endl;return 0;
}

#include <cstdio>
#include <iostream>
#include <string>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <map>
#include <stack>
using namespace std;
stack<int>kepa;
int main()
{int n,m,j,a;int t=1;string s;cin>>n>>m;kepa.push(0);for(int i=1;i<=m;i++){int it=kepa.top();int res=1;int temp=0;cin>>s;if(s[0]=='u'){cin>>a;while(a--){kepa.pop();}}else{if(s[0]=='-'){res=-1;}for(j=0;j<s.length();j++){if(j==0&&res==-1){continue;}temp=temp*10+s[j]-'0';}temp=temp*res;it+=temp;while(it<0)it+=n;it%=n;kepa.push(it);}}cout<<kepa.top()<<endl;return 0;
}

#include <cstdio>
#include <iostream>
#include <string>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <map>
#include <stack>
using namespace std;
int main()
{int n,m;int t=1;int num[1000];num[0]=0;string s;int a,b;int sw;cin>>n>>m;for(int i=1;i<=m;i++){cin>>s;if(s[0]=='u'){cin>>a;t-=a+1;}else{a=atoi(s.c_str());sw=num[t-1];sw+=a;sw%=n;if(sw<0) sw+=n;num[t]=sw;}t++;}cout<<num[t-1]<<endl;return 0;
}

#include <iostream>
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
using namespace std;
int dp[105][105],w[105],pre[105];
int main()
{
int n,m,i,j;
cin>>n>>m;
for(i=1;i<=n;i++)cin>>w[i];
pre[0]=m;
for(i=1;i<=n;i++)
{pre[i]=pre[i-1]*2/3;
}
memset(dp,-1,sizeof(dp));
dp[0][0]=0;
for(i=1;i<=n;i++)
{dp[i][0]=min(w[i],pre[0]);
}
for(i=0;i<=n;i++)
{for(j=0;j<=n;j++){if(dp[i][j]==-1)continue;dp[i+1][j+1]=max(dp[i+1][j+1],dp[i][j]+min(w[i+1],pre[j+1]));dp[i+2][j]=max(dp[i+2][j],dp[i][j]+min(w[i+2],pre[j]));dp[i+3][0]=max(dp[i+3][0],dp[i][j]+min(w[i+3],pre[0]));}
}
int ans=0;for(j=0;j<=n;j++){ans=max(ans,dp[n][j]);}cout<<ans<<endl;
return 0;
}