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 <bits/stdc++.h>
using namespace std;
int A[8][8], cnt=0;
bool B[8][8];
void put(int r,int c,int k){
	cnt += k;
	B[r][c] += k;
	for(int i=0; i<8; i++)
		A[r][i] += k,
		A[i][c] += k;
	A[r][c] -= k;
	for(int i=1; r+i<8 && c+i<8; i++)
		A[r+i][c+i] += k;
	for(int i=1; r-i>=0 && c-i>=0; i++)
		A[r-i][c-i] += k;
	for(int i=1; r+i<8 && c-i>=0; i++)
		A[r+i][c-i] += k;
	for(int i=1; r-i>=0 && c+i<8; i++)
		A[r-i][c+i] += k;
}
bool dfs(int r){
	if(cnt == 8) return 1;
	for(int i=r; i<8; i++)
		for(int j=0; j<8; j++){
			if(A[i][j]) continue;
			put(i,j,1);
			if(dfs(i+1)) return 1;
			put(i,j,-1);
		}
	return 0;
}
int main()
{	
	memset(A,0,sizeof(A));
	memset(B,0,sizeof(B));
	int n, r, c; 
	cin >> n;
	while(n--){
		cin >> r >> c;
		put(r,c,1);
	}
	dfs(0);
	for(int i=0; i<8; i++){
		for(int j=0; j<8; j++)
			cout << (B[i][j]?'Q':'.');
		cout << endl;
	}
	return 0;
}
