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
#define FREE -1
#define NOTFREE 1

int row[N],col[N],dpos[2*N-1],dneg[2*N-1],map[N][N];

void init();
void print();
void rec(int);

int main(){
  init();
  int r,c,n,i;
  scanf("%d",&n);
  for(i=0;i<n;i++){
    scanf("%d%d",&r,&c);
    map[r][c]=1;
  }
  rec(0);
  return 0;
}

void init(){
  int i;
  for(i=0;i<N;i++){
    row[i]=FREE;
    col[i]=FREE;
    dpos[i]=FREE;
    dneg[i]=FREE;
  }
  for(;i<2*N-1;i++){
    dpos[i]=FREE;
    dneg[i]=FREE;
  }
}

void print(){
  int i,j;
  for(i=0;i<N;i++){
    for(j=0;j<N;j++){
      if(map[i][j]==1){
	if(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 rec(int i){
  int j;
  if(i==N){
    print();
    return;
  }
  for(j=0;j<N;j++){
    if(NOTFREE==col[j]||
       NOTFREE==dpos[i+j]||
       NOTFREE==dneg[i-j+N-1])continue;
    row[i]=j;
    col[j]=dpos[i+j]=dneg[i-j+N-1]=NOTFREE;
    rec(i+1);
    row[i]=col[j]=dpos[i+j]=dneg[i-j+N-1]=FREE;
  }
}

