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<cstdio>
#include<algorithm>
using namespace std;
const int MAXN=8+4,
          MAXM=16+4;
char a[MAXN][MAXN];
int zuoxie[MAXM],youxie[MAXM],lie[MAXN],pos[MAXN];
int r[MAXN],c[MAXN];
int n;
void wwq(){
	int cnt=0;
	for(int i=1;i<=n;i++)
		if(a[r[i]][c[i]]=='Q') cnt++;
	if(cnt==n){
		for(int i=0;i<8;i++){
			for(int j=0;j<8;j++) cout<<a[i][j];
			cout<<endl;
		}
		//cout<<endl;
	}
	
}
void dfs(int i){
	for(int j=0;j<8;j++){
		if(zuoxie[i+j-1]==0 && youxie[i-j+7]==0 && lie[j]==0){
			zuoxie[i+j-1]=1;
			lie[j]=1;
			youxie[i-j+7]=1;
			a[i][j]='Q';
			if(i==7) wwq();
			else dfs(i+1);
			zuoxie[i+j-1]=0;
			youxie[i-j+7]=0;
			lie[j]=0;
			a[i][j]='.';
		}
	}
}
int main(){
	//freopen("equeen.out","w",stdout);
    for(int i=0;i<8;i++)
		for(int j=0;j<8;j++) a[i][j]='.';
    cin>>n;
    for(int i=1;i<=n;i++) cin>>r[i]>>c[i];
	dfs(0);
	return 0;
} 