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 <stdio.h>
#include <bitset>
using namespace std;
const int N=8;
char c[N][N];
bitset<N*10> u,d,l,r;
bool dfs(int now){
    if(now>=N)return true;
    if(d[now])return dfs(now+1);
    for(int i=0;i<N;i++)if(!u[i]&&!l[i+now]&&!r[i-now+10]){
        c[now][i]='Q';
        u[i]=l[i+now]=r[i-now+10]=true;
        if(dfs(now+1))return true;
        u[i]=l[i+now]=r[i-now+10]=false;
        c[now][i]='.';
    }
    return false;
}
int main(){
    int n,x,y;
    scanf("%d",&n);
    for(int i=0;i<N;i++)for(int j=0;j<N;j++)c[i][j]='.';
    u.reset();
    r.reset();
    d.reset();
    l.reset();
    while(n--){
        scanf("%d%d",&x,&y);
        c[x][y]='Q';
        d[x]=u[y]=l[x+y]=r[y-x+10]=true;
    }
    dfs(0);
    for(int i=0;i<N;i++){
        for(int j=0;j<N;j++)printf("%c",c[i][j]);
        printf("\n");
    }
}
