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>
#define rep(i,a,b) for(int i=int(a);i<int(b);++i)
#define SIZE 200005
#define INF 1000000005LL
#define MOD 1000000007

using namespace std;
typedef long long int ll;
typedef pair <int,int> P;

int k;
vector<P> rc;
int u[10];

int main(){

  cin >> k;
  rep(i,0,8) u[i] = i;
  rep(i,0,k){
    int b,c;
    cin >> b >> c;
    rc.push_back(P(b,c));
  }

  do{
    bool a = true;
    rep(i,0,k) a &= (u[rc[i].first] == rc[i].second);
    if(!a) continue;

    rep(i,0,8) rep(j,0,8) if(i != j){
      if(abs(i-j) == abs(u[i]-u[j])) a = false;
    }
    if(a){
      rep(i,0,8){
        rep(j,0,8){
          cout << (u[i]==j ? "Q" : ".") ;
        }
        cout << endl;
      }
      break;
    }

  }while(next_permutation(u,u+8));

  return 0;
}

