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<cstring>
using namespace std;
char map[8][8];

bool check(int x,int y){
    int dx[]={-1, -1, -1, 1, 1, 1, 0, 0};
    int dy[]={ 0, 1, -1, 0, 1, -1, 1, -1};
    for(int i=0;i<8;i++){
        for(int j=1;;j++){
            int new_x=x+dx[i]*j;
            int new_y=y+dy[i]*j;
            if(new_x<0 || new_y<0 || new_x>=8 || new_y>=8)
                break;
            if(map[new_x][new_y]=='Q')
                return false;
        }
    }
    return true;
}
void dfs(int row){
    if(row>=8){
        for(int i=0;i<8;i++){
            for(int j=0;j<8;j++)
                cout<<map[i][j];
            cout<<endl;
        }
        return;
    }
    for(int i = 0; i < 8; i++)
        if(map[row][i] == 'Q'){
           dfs(row + 1);
           return;
        }
             
    for(int col=0;col<8;col++){
        if(check(row,col)){
            map[row][col]='Q';
            dfs(row+1);
            map[row][col]='.';
        }
    }
}
int main(){

    int n,x,y;cin>>n;
    for(int i=0;i<8;i++)
        for(int j=0;j<8;j++)
            map[i][j]='.';
    while(n--){
        cin>>x>>y;
        map[x][y]='Q';
    }
    dfs(0);
    
}
