/**
    题意：给一个图，然后判断是哪一个图在上边，并且该框是>=3*3
    做法：模拟
**/
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <cmath>
#include <algorithm>
#include <queue>
#define maxn 110
using namespace std;
int vis[maxn][maxn];
int dx[4] = {0,0,-1,1};
int dy[4] = {1,-1,0,0};
char ch[maxn][maxn];
int alp[30];
struct Node
{
    int x;
    int y;
} node[maxn];
struct NN
{
    int x[5];
    int y[5];
    char c[10];
} no[30];
int n,m;
int check(int x,int y)
{
    if(x >=0 && x < n && y >= 0 && y < m && vis[x][y] == 0) return 1;
    return 0;
}
int bfs(int x,int y)
{
    queue<Node>que;
    Node tmp,now;
    vis[x][y] = 1;
    tmp.x = x;
    tmp.y = y;
    char c = ch[x][y];
    que.push(tmp);
    int sum = 1;
    while(!que.empty())
    {
        now = que.front();
        que.pop();
        for(int i=0; i<4; i++)
        {
            tmp.x = now.x + dx[i];
            tmp.y = now.y + dy[i];
            if(check(tmp.x,tmp.y) && ch[tmp.x][tmp.y] == c)
            {
                vis[tmp.x][tmp.y] = 1;
                que.push(tmp);
                sum++;
            }
        }
    }
    return sum;
}
int solve()
{
    for(int i=0; i<n; i++)
    {
        for(int j=0; j<m; j++)
        {
            int num1 = 0;
            int num2 = 0;
            int num = 0;
            int res = 0;
            if(ch[i][j] != ‘.‘ && vis[i][j] == 0)
            {
                char c = ch[i][j];
                for(int ii = i; ii<n; ii++)
                {
                    if(ch[ii][j] == c) num1++;
                    else break;
                }
                for(int jj= j; jj<m; jj++)
                {
                    if(ch[i][jj] == c) num2++;
                    else break;
                }
                num = bfs(i,j);
                if(num == 2*( num1 + num2 -2) && num1 >=3 && num2 >= 3 && num >= 8)
                {
                    int tt = ch[i][j] - ‘A‘;
                    alp[tt] = 1;
                    no[tt].x[0] = i;
                    no[tt].x[1] = i+ num1 -1;
                    no[tt].y[0] = j;
                    no[tt].y[1] = j + num2 -1;
                }
            }
        }
    }
    for(int i=0; i<26; i++)
    {
        if(alp[i] == 1)
            for(int j=0; j<26; j++)
            {
                if(alp[j] == 1 && no[i].x[0] < no[j].x[0] && no[i].x[1] > no[j].x[1] && no[i].y[0]< no[j].y[0] && no[i].y[1] > no[j].y[1])
                {
                    alp[i] = 0;
                   // break;
                }
            }
    }
    for(int i=0; i<26; i++)
    {
        if(alp[i])
        {
            printf("%c",i+‘A‘);
        }
    }
    printf("\n");
}
int main()
{
  //  freopen("in.txt","r",stdin);
    while(~scanf("%d %d",&n,&m))
    {
        if(n == 0 && m == 0) break;
        for(int i=0; i<n; i++)
        {

            scanf("%s",ch[i]);

        }
        memset(vis,0,sizeof(vis));
        memset(alp,0,sizeof(alp));
        solve();
    }
}