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<cstdio>
#include<cstring>
#include<iostream>
#define N 8
#define Free 1
#define NotFree 0
using namespace std;
int row[N], col[N], leftd[2*N-1], rightd[2*N-1];
bool M[8][8];

void output()
{
  for(int i=0; i<N; i++)
    for(int j=0; j<N; j++)
  {
    if(M[i][j])
    {
      if(row[i] == j) continue;
      else return ;
    }
  }
  for(int i=0; i<N; i++)
  {
    for(int j=0; j<N; j++)
      cout << ((row[i] == j) ? "Q" : ".") ;
    cout << endl;
  }
}

void dfs(int u)
{
  if(u == N)
  {
    output(); return ;
  }
  for(int j=0; j<N; j++)
  {
    if(col[j]==Free && leftd[j+u] == Free && rightd[u-j + N-1] == Free)
      { row[u] = j; col[j] = leftd[j+u] = rightd[u-j+N-1] = NotFree;  }
    else continue;
    //int v = u;
    dfs(u+1);
    row[u] = col[j] = leftd[j+u] = rightd[u-j+N-1] = Free;
  }

}
int main()
{
  memset(M, 0, sizeof(M));
  int k;
  scanf("%d", &k);
  for(int i=0; i<k; i++)
  {
    int r, c;
    scanf("%d%d", &r, &c);
    M[r][c] = true;
  }
  for(int i=0; i<N; i++)
  {
    row[i] = col[i] = Free;
  }
  for(int i=0; i<2*N-1; i++)
  {
    leftd[i] = rightd[i] = Free;
  }
  dfs(0);
  return 0;
}

