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 <stdio.h>
#define N 8
int row[N]={},col[N]={},dpos[2*N-1]={},dneg[2*N-1]={};
int board[N][N];
void printBoard(){
  int i,j;
  for(i=0;i<N;i++){
    for(j=0;j<N;j++){
      if(board[i][j]==1&&row[i]!=j)return;
    }
  }
  for(i=0;i<N;i++){
    for(j=0;j<N;j++){
      if(row[i]==j)printf("Q");
      else printf(".");
    }
    printf("\n");
  }
}
void putQueen(int i){
  int j;
  if(i==N){
    printBoard();
    return;
  }
  for(j=0;j<N;j++){
    if(col[j]==1||dpos[i+j]==1||dneg[i-j+N-1]==1)continue;
    row[i]=j;
    col[j]=dpos[i+j]=dneg[i-j+N-1]=1;
    putQueen(i+1);
    row[i]=col[j]=dpos[i+j]=dneg[i-j+N-1]=0;
  }
}

int main(){
  int k,r,c,i,j;
  
  for(i=0;i<N;i++){
    for(j=0;j<N;j++){
      board[i][j]=0;
    }
  }
  
  scanf("%d",&k);
  for(i=0;i<k;i++){
    scanf("%d %d",&r,&c);
    board[r][c]=1;
  }
  putQueen(0);
  return 0;
}