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<cassert>
using namespace std;

#define N 8
#define F -1
#define NF 1

int str[N], bor[N], dp[2*N-1], dn[2*N-1];

bool X[N][N];

void ini(){
  int i;
  for(i=0;i<N;i++){
    str[i]=F;
    bor[i]=F;
  }
  for(i=0;i<2*N-1;i++){
    dp[i]=F;
    dn[i]=F;
  }
}

void prb(){
  int i, j;
  for(i=0;i<N;i++){
    for(j=0;j<N;j++){
      if(X[i][j]){
        if(str[i]!=j) return;
      }
    }
  }
  for(i=0;i<N;i++){
    for(j=0;j<N;j++){
      cout << ((str[i]==j)?"Q" : ".");
    }
    cout << endl;
  }
}

void rec(int i){
  int j;
  if(i==N){
    prb();
    return;
  }

  for(j=0;j<N;j++){
    if(NF==bor[j] || NF==dp[i+j] || NF==dn[i-j+N-1]) continue;
    str[i]=j;bor[j]=dp[i+j]=dn[i-j+N-1]=NF;
    rec(i+1);
    str[i]=bor[j]=dp[i+j]=dn[i-j+N-1]=F;
  }
}

int main(){
  int i, j, k, r, c;
  ini();

  for(i=0;i<N;i++)
    for(j=0;j<N;j++)X[i][j]=false;

  cin>>k;
  for(i=0;i<k;i++){
    cin >> r >> c;
    X[r][c]=true;
  }

  rec(0);

  return 0;
}
