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;

constexpr int N = 8;

int grid[8][8];
int row[N], col[N];
int dpos[2 * N - 1], dneg[2 * N - 1];

void Print(){
  for( int i = 0; i < N; i++ ){
    for( int j = 0; j < N; j++ ){
      if( grid[i][j] && row[i] != j ) return;
    }
  }

  for( int i = 0; i < N; i++ ){
    for( int j = 0; j < N; j++ ){
      if( row[i] == j ) cout << "Q";
      else cout << ".";
    }
    cout << endl;
  }  
}

void dfs( int i ){
  if( i == N ){
    return Print();
  }

  for( int j = 0; j < N; j++ ){
    // 0:OK, 1:NG
    if( col[j] || dpos[i + j] || dneg[i - j + ( N - 1 )] ){
      continue;
    }

    row[i] = j;
    col[j] = dpos[i + j] = dneg[i - j + ( N - 1 )] = 1;
    dfs( i + 1 );
    col[j] = dpos[i + j] = dneg[i - j + ( N - 1 )] = 0;
  }
}

int main(){  
  int k; cin >> k;
  for( int i = 0; i < k; i++ ){
    int r, c; cin >> r >> c;
    grid[r][c] = 1;
  }

  dfs( 0 );
}

