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, r

const 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;
}