依次记录1的数量x和0的数量y,每一个位置判断1或0，因为是回文，所以只要第一次遇到与开头相同的数就成立，x与y
均大于2则一定有一条路使其为回文。因为结尾不为起始点，n与m为1的情况排除
#include<bits/stdc++.h>
usingnamespacestd;
  
intmain()
{
    intT;
    cin>>T;
    while(T--)
    {
        longlongintn,m,x=0,y=0;
         cin>>n>>m;
        longlonginta[n][m];
        for(longlonginti=0;i<n;i++)
        {
            for(longlongintj=0;j<m;j++)
            {
               scanf("%1lld",&a[i][j]);
               if(a[i][j]==1)
               {x++;}
               else
               {y++;}
            }
            scanf("\n");
        }
        for(longlonginti=0;i<n;i++)
        {
            for(longlongintj=0;j<m;j++)
            {
                if(a[i][j]==1&&x>=2)
                {cout<<"Y";}
                elseif(a[i][j]==0&&y>=2)
                {cout<<"Y";}
                else
                {cout<<"N";}
            }
            cout<<'\n';
        }
         
    }
}