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 <stdio.h>
#include <string.h>
 
#define N 8
 
int a[N][N] = {{}};
int row[N] = {};
int l[15] = {};
int r[15] = {};
int s[N] = {};
 
void print(void);
void backtrack(int); 
 
int main(){
 
 
  int i,j,r,c,k;
 
 
  scanf("%d",&k);
 
  for(i = 0;i < k;i++){
    scanf("%d%d",&r,&c);
    a[r][c] = 1;
  }
  backtrack(0);
}
 
void backtrack(int i){
  int j;
 
  if(i == N){
    print();
    return;
  }
 
  for(j = 0;j < N;j++){
    if(s[j] == 1 || l[i + j] == 1 || r[i - j + N - 1] == 1) continue;
 
    row[i] = j;
    l[i + j] = r[i - j + N - 1] = s[j] = 1;
 
    backtrack(i + 1);
 
    row[i] = s[j] = l[i + j] = r[i - j + N - 1] = 0;
 
  }
 
}
 
void print(){
 
  int i,j;
 
  for(i = 0;i < N;i++){
    for(j = 0;j < N;j++){
      if(a[i][j] == 1){
        if(row[i] != j) {
          return;
        }
      }
    }
  }
  for(i = 0;i < N;i++){
    for(j = 0;j < N;j++){
      if(row[i] == j) printf("Q");
      else printf(".");
    }
    printf("\n");
  }
 
}
