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 a[8]={},b[8]={},c[15]={},d[15]={},ans[8][8]={};
void print(void);
void dfs(int);
 
int main(){
  int n,r,c1;
  cin>>n;
  for(int i=0;i<n;i++){
    cin>>r>>c1;
    ans[r][c1]=1;
  }
  dfs(0);
  return 0;
}
void dfs(int x){
  if(x==8){
  for(int i=0;i<8;i++)
    for(int j=0;j<8;j++)
      if(ans[i][j]==1)
	if(a[i]!=j)return;
 
  for(int i=0;i<8;i++){
    for(int j=0;j<8;j++)
      if(a[i] == j)printf("Q");
      else printf(".");
    printf("\n");
  }
    return;
  }
 
  for(int i=0;i<8;i++){
    if(b[i]==1||c[x+i]==1||d[x-i+7]==1)continue;
    a[x]=i;
    b[i]=c[x+i]=d[x-i+7]=1;
    dfs(x+1);
    a[x]=b[i]=c[x+i]=d[x-i+7]=0;
  }
}