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 MAX 8
#define N 8
#define NOT_FREE -1
#define FREE 1
using namespace std;
int col[MAX],dpos[2*MAX-1],dneg[2*MAX-1],row[MAX];
int X[MAX][MAX];

void printBoard(){
    int i,j;
    for(i=0;i<N;i++) {
		for (j=0;j<N;j++) {
			if(X[i][j]!=FREE){
				if(row[i]!=j)return;
			}
		}
	}
    for(i =0;i<N;i++){
        for(j=0;j<N;j++){
            if(row[i]==j)cout<<'Q';
            else cout<<".";
        }
        cout<<endl;
    }
}

void putQueen(int i){
    int j;
    if(i==N){
        printBoard();
        return;
    }
    for(j=0;j<N;j++){
        if(col[j]==NOT_FREE||dpos[i+j]==NOT_FREE||dneg[i-j+N-1]==NOT_FREE)continue;
        row[i]=j;
        col[j]=dpos[i+j]=dneg[i-j+N-1]=NOT_FREE;
        putQueen(i+1);
        row[i]=col[j]=dpos[i+j]=dneg[i-j+N-1]=FREE;
    }
}

int main(){
    int a,b,n,i,j;
    for(i=0;i<N*2-1;i++){
        dpos[i]=FREE;
        dneg[i]=FREE;
    }
    for(i=0;i<MAX;i++){
        col[i]=FREE;
        row[i]=FREE;
    }
    for(i=0;i<MAX;i++){
        for(j=0;j<MAX;j++){
            X[i][j]=FREE;
        }
    }
    cin>>n;
    for(i=0;i<n;i++){
        cin>>a>>b;
        X[a][b]=NOT_FREE;
        // row[b]=a;
        // col[a]=b;
    }
    putQueen(0);
}


