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;

typedef long long ll;
#define endl '\n'

const int chessBoardSize = 8;
#define n chessBoardSize

int col[n], dig1[2*n], dig2[2*n];
int sol[n];
int ans[n];
bool exist[n];
bool found = false;

void search(int x){

    if(x==n){
        found=true;
        return;
    }

    if(exist[x]){
        search(x+1);
        return;
    }
    
    for(int y=0;y<n;y++){
        if(col[y] || dig1[y+x] || dig2[y-x+n-1])
            continue;
        sol[x]=y;
        col[y] = dig1[y+x] = dig2[y-x+n-1] = true;
        search(x+1);
        if(found)
            return;
        col[y] = dig1[y+x] = dig2[y-x+n-1] = false;
    }

}
int main() {
   
    std::ios::sync_with_stdio(false);

    int t, x, y;
    cin>>t;
    while(t--){
        cin>>x>>y;
        sol[x]=y;
        col[y] = dig1[y+x] = dig2[y-x+n-1] = true;
        exist[x]=true;
    }

    search(0);
    
    for(int i=0;i<n;i++){
        for(int j=0;j<n;j++){
            if(j!=sol[i])
                cout<<'.';
            else
                cout<<'Q';
        }
        cout<<endl;
    }
    
    return 0;
}

