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;

bool usedA[8];
bool usedB[8];
bool usedC[16];
bool usedD[16];

char t[8][8];


bool dfs(int y,int x,int cnt){
  if(y==8){
    return (cnt==8);
  }
  if(x==8){
    return dfs(y+1,0,cnt);
  }
  if( usedA[y] || usedB[x] || usedC[y+x] || usedD[y-x+8] ){
    return dfs(y,x+1,cnt);
  }
  t[y][x]='Q';
  usedA[y]=usedB[x]=usedC[y+x]=usedD[y-x+8]=true;
  if( dfs(y,x+1,cnt+1) )return true;
  usedA[y]=usedB[x]=usedC[y+x]=usedD[y-x+8]=false;
  t[y][x]='.';
  if( dfs(y,x+1,cnt) )return true;
  return false;
}

int main(){
  for(int i=0;i<8;i++)
    for(int j=0;j<8;j++)
      t[i][j]='.';

  int n;
  cin>>n;
  for(int i=0;i<n;i++){
    int y,x;
    cin>>y>>x;
    t[y][x]='Q';
    usedA[y]=usedB[x]=usedC[y+x]=usedD[y-x+8]=true;
  }
  dfs(0,0,n);
  for(int i=0;i<8;i++){
    for(int j=0;j<8;j++){
      cout<<t[i][j];
    }
    cout<<endl;
  }
  return 0;
}

