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...
//By Vlgd
#include<bits/stdc++.h>
using namespace std;
const int N=8;
int borad[N][N];
int r[N],c[N]={0},pd[2*N-1]={0},nd[2*N-1]={0};
void dfs(int x){
	if(x==N){
		for(int i=0;i<N;i++){
			for(int j=0;j<N;j++){
				if(j==r[i]) printf("Q");
				else printf(".");
			}
			printf("\n");
		}
		return;
	}
	if(r[x]!=-1) dfs(x+1);
	else {
		for(int y=0;y<N;y++){
			if(c[y]==0 && pd[x+y]==0 && nd[x-y+N-1]==0){
				r[x]=y;c[y]=1;pd[x+y]=1;nd[x-y+N-1]=1;
				dfs(x+1);
				r[x]=-1;c[y]=0;pd[x+y]=0;nd[x-y+N-1]=0;
			}
		}
	}
}
int main(){
	int k;scanf("%d",&k);
	int x,y;
	memset(r,-1,sizeof(r));
	for(int i=0;i<k;i++){
		scanf("%d%d",&x,&y);
		r[x]=y;c[y]=1;pd[x+y]=1;nd[x-y+N-1]=1;
	}
	dfs(0);
	return 0;
}
/*
2
2 2
5 3
*/
