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>
#define rep(i, n) for(int i = 0; i < (n); i++)
using namespace std;
using ll = long long;
using pii = pair<int, int>;

int main() {
	int k;
	cin >> k;
	vector<int> r(k), c(k);
	vector<int> rf(8), cf(8);
	vector<vector<char>> ans(8, vector<char>(8, '.'));
	rep(i, k) {
		cin >> r[i] >> c[i];
		rf[r[i]] = 1;
		cf[c[i]] = 1;
		ans[r[i]][c[i]] = 'Q';
	}
	vector<int> vr, vc;
	rep(i, 8) {
		if (!rf[i]) vr.push_back(i);
		if (!cf[i]) vc.push_back(i);
	}
	sort(vr.begin(), vr.end());
	sort(vc.begin(), vc.end());
	do {
		do {
			bool ok = true;
			rep(i, 8 - k) rep(j, k) {
				if (abs(vr[i] - r[j]) == abs(vc[i] - c[j])) ok = false;
			}
			rep(i, 8 - k) rep(j, 8 - k) {
				if (i == j) continue;
				if (abs(vr[i] - vr[j]) == abs(vc[i] - vc[j])) ok = false;
			}
			if (ok) {
				rep(i, 8 - k) ans[vr[i]][vc[i]] = 'Q';
			}
		} while (next_permutation(vc.begin(), vc.end()));
	} while (next_permutation(vr.begin(), vr.end()));
	rep(i, 8) {
		rep(j, 8) cout << ans[i][j];
		cout << endl;
	}
	return 0;
}
