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 <algorithm>
#include <vector>
#include <set>
#include <cmath>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
const ll INF = -1e18;
const int I=2001;
int ip() {int x; cin>>x; return x;}

int main(){
  int k; cin>>k;
  int a[k], b[k]; for(int i=0;i<k;i++) a[i]=ip(), b[i]=ip();
  int p[8]={0,1,2,3,4,5,6,7};
  do {
    bool ok=true;
    for(int i=0;i<k;i++)
      if (p[a[i]]!=b[i]) ok=false;
    if (!ok) continue;
    bool mp[8][8]; fill(mp[0], mp[8], false);
    for(int i=0;i<8&&ok;i++) {
      if (mp[i][p[i]]) ok=false;
      else {
        int x[4]={1,1,-1,-1}, y[4]={1,-1,1,-1};
        for(int j=0;j<4;j++) {
          int sx=i+x[j], sy=p[i]+y[j];
          while(0 <= sx && sx < 8 && 0 <= sy && sy < 8) {
            mp[sx][sy]=true;
            sx+=x[j], sy+=y[j];
          }
        }
      }
    }
    if (ok) {
      for(int i=0;i<8;i++) {
        for(int j=0;j<8;j++) {
          if (p[i] == j) cout<<"Q"; else cout<<".";
        } cout<<endl;
      }
      break;
    }
  } while(next_permutation(p,p+8));
}
