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...
#define _CRT_SECURE_NO_WARNINGS

#include<iostream>
#include<functional>
#include<algorithm>
#include<stack>
#include<queue>
#include<iomanip>
#include<set>
#include<map>
#include<utility>
#include<string>
#include<vector>

using namespace std;

#define LL long long
const int dx[4] = { 1,-1,0,0 };
const int dy[4] = { 0,0,1,-1 };
bool OOA(int x, int y, int H, int W) { return (0 <= x && x<H) && (0 <= y && y<W); }

int k;
int Q[8];
pair<int,int>fq[8];

int main() {
	cin>>k;
	for(int i=0;i<8;i++)Q[i]=-1;
	for(int i=0;i<k;i++){
		int r,c;
		cin>>r>>c;
		fq[i].first=r;
		fq[i].second=c;
		Q[r]=c;
	}
	int rc[8]={0,1,2,3,4,5,6,7};
	while(next_permutation(rc,rc+8)){
		bool flg=false;
		for(int i=0;i<k;i++){
			flg|=(rc[fq[i].first]!=fq[i].second);
			if(flg)break;
		}
		if(flg)continue;
		set<int>dd,ud;
		for(int i=0;i<8;i++){
			dd.insert(i-rc[i]);
			ud.insert(i+rc[i]);
		}
		if(dd.size()+ud.size()!=16)continue;
		for(int i=0;i<8;i++){
			for(int j=0;j<8;j++){
				if(rc[i]==j)cout<<'Q';
				else cout<<'.';
			}
			cout<<endl;
		}
		break;
	}

	return 0;
}

