CodeForces 90B African Crossword【模拟】

African Crossword

Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 90B

Description

An African crossword is a rectangular table n × m in size. Each cell of the table contains exactly one letter. This table (it is also referred to as grid) contains some encrypted word that needs to be decoded.

To solve the crossword you should cross out all repeated letters in rows and columns. In other words, a letter should only be crossed out if and only if the corresponding column or row contains at least one more letter that is exactly the same. Besides, all such letters are crossed out simultaneously.

When all repeated letters have been crossed out, we should write the remaining letters in a string. The letters that occupy a higher position follow before the letters that occupy a lower position. If the letters are located in one row, then the letter to the left goes first. The resulting word is the answer to the problem.

You are suggested to solve an African crossword and print the word encrypted there.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 100). Next n lines contain m lowercase Latin letters each. That is the crossword grid.

Output

Print the encrypted word on a single line. It is guaranteed that the answer consists of at least one letter.

Sample Input

Input

3 3
cba
bcd
cbc

Output

abcd

Input

5 5
fcofd
ooedo
afaoa
rdcdf
eofsf

Output

codeforces

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<climits>
#include<string>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<algorithm>
using namespace std;
#define rep(i,j,k)for(i=j;i<k;i++)
#define per(i,j,k)for(i=j;i>k;i--)
#define MS(x,y)memset(x,y,sizeof(x))
#define max(a,b) a>b?a:b
#define min(a,b) a<b?a:b
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
typedef long long LL;
const int INF=0x7ffffff;
#define lson rt<<1, l, m
#define rson rt<<1|1, m+1, rconst int M=100+1;
char mp[M][M];
int flag[M][M];
int i,j,k,n,m;void cheak(int nn,int mm)
{for(int k=nn+1;k<n;k++){if(mp[k][mm]==mp[nn][mm])flag[k][mm]++,flag[nn][mm]++;}for(int k=mm+1;k<m;k++){if(mp[nn][k]==mp[nn][mm])flag[nn][k]++,flag[nn][mm]++;}return ;
}int main()
{scanf("%d%d",&n,&m);MS(flag,0);for(int i=0;i<n;i++)scanf("%s",mp[i]);for(int i=0;i<n;i++)for(int j=0;j<m;j++){cheak(i,j);}for(i=0;i<n;i++)for(j=0;j<m;j++)if(!flag[i][j])printf("%c",mp[i][j]);printf("\n");return 0;
}

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