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 <iostream>
#include <bits/stdc++.h>
using namespace std;

int s[8];
char a[10][10]={"........","........","........","........","........","........","........","........"};
void dfs(int n,int k);
void dfs(int n,int k)
{
   int s1[8],b[10][10];
   for(int i=0;i<8;i++)
      s1[i]=s[i];
      for(int i=0;i<10;i++)
         for(int j=0;j<10;j++)
         b[i][j]=a[i][j];
   if(n==8)
   {
      for(int i=0;i<8;i++)
         puts(a[i]);
         return;
   }
   if(s[k]>=0)
   {
      dfs(n,k+1);
      return;
   }
   for(int i=0;i<8;i++)
   {
      int f=1;
      for(int j=0;j<8;j++)
      {
         if(s[j]==i||(s[j]>=0&&fabs(s[j]-i)==fabs(j-k)))
         {
            f=0;break;
         }
      }
      if(f)
      {
         s[k]=i;
         a[k][i]='Q';
         dfs(n+1,k+1);
         for(int i=0;i<8;i++)
            s[i]=s1[i];
             for(int i=0;i<10;i++)
         for(int j=0;j<10;j++)
         a[i][j]=b[i][j];
      }
   }
}
int main()
{
    int x,y,t;
    cin>>t;
    memset(s,-1,sizeof(s));
    for(int i=0;i<t;i++)
    {
       cin>>y>>x;
       a[y][x]='Q';
       s[y]=x;
    }
    dfs(t,0);
    return 0;
}

