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<cassert>
using namespace std;
#define N 8
#define FREE -1
#define NOT_FREE 1
int row[N], col [N], dpos[2 * N - 1], dneg[2 * N - 1];
bool X[N][N];
void initialize() {
for(int i = 0; i < N; i++ ){row[i] = FREE, col[i] = FREE; }
for(int i=0; i<2* N - 1; i++ ){ dpos[i] = FREE; dneg[i] = FREE;}
}
void printBoard(){
for (int i = 0; i< N; i++ ){
for (int j = 0; j < N; j++ ){
if (X[i][j]){
if (row[i] != j)return;
}
}
}
for (int i = 0; i < N; i++ ){
for (int j = 0; j < N; j++ ){ cout << ((row[i] == j)? "Q":".");}
cout << endl;
}
}
void recursive(int i){
if (i == N){ // クイーンの配置に成功
printBoard(); return;
}
for(int j = 0; j< N; j++ ){
// (i,j)が他のクイーンに襲撃されている場合は無視
if( NOT_FREE == col[j] ||
NOT_FREE == dpos[i + j] ||
NOT_FREE == dneg[i - j + N - 1] ) continue;
//(i,j)にクイーンを配置する
row[i] = j; col[j] = dpos[i + j] = dneg[i - j + N - 1] = NOT_FREE;
// 次の行を試す
recursive(i + 1);
// (i, j)に配置されているクイーンを取り除く
row[i] = col[j] = dpos[i + j] = dneg[i - j + N - 1] = FREE;
// クイーンの配置に失敗
}
}
int main(){
initialize();
for (int i = 0; i < N; i++)
for (int j = 0; j < N; j++ ) X[i][j] = false;
int k; cin >> k;
for (int i = 0; i<k; i++ ){
int r, c; cin >>r >> c;
X[r][c] = true;
}
recursive(0);
return 0;    
}


