The goal of 8 Queens Problem is to put eight queens on a chess-board such that none of them threatens any of others. A queen threatens the squares in the same row, in the same column, or on the same diagonals as shown in the following figure.

<image>


For a given chess board where $k$ queens are already placed, find the solution of the 8 queens problem.

Constraints

* There is exactly one solution

Input

In the first line, an integer $k$ is given. In the following $k$ lines, each square where a queen is already placed is given by two integers $r$ and $c$. $r$ and $c$ respectively denotes the row number and the column number. The row/column numbers start with 0.

Output

Print a $8 \times 8$ chess board by strings where a square with a queen is represented by 'Q' and an empty square is represented by '.'.

Example

Input

2
2 2
5 3


Output

......Q.
Q.......
..Q.....
.......Q
.....Q..
...Q....
.Q......
....Q...
#include<bits/stdc++.h>
using namespace std;
   
int n,a,b,A[8],B[8],C[33],D[33],M[8][8],IN[8];

void dfs(int x){

  if(x==8){
    for(int i=0;i<8;i++){
      for(int j=0;j<8;j++)
        if(M[i][j])cout<<'Q';
        else cout<<'.';
      cout<<endl;
    }
    exit(0);
  }

  for(int j=0;j<8;j++){
    if( (IN[x]&&j!=IN[x]-1) ||B[j]==1||C[x+j]==1||D[x-j+10]==1)continue;
    M[x][j]=A[x]=B[j]=C[x+j]=D[x-j+10]=1;
    dfs(x+1);
    M[x][j]=A[x]=B[j]=C[x+j]=D[x-j+10]=0;
  }
   

}
   
int main(){
  cin>>n;
  while(n--){
    cin>>a>>b;
    M[a][b]=1;
    IN[a]=b+1;
  }
  dfs(0);
}
