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;
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,n)   FOR(i,0,n)
#define pb emplace_back
typedef long long ll;
typedef pair<int,int> pint;

int x[8],y[8],ax[8],ay[8];
bool used[8];
void dfs(int cur){
    if(cur==8){
        rep(i,8) ax[i]=x[i],ay[i]=y[i];
        return;
    }
    int ty=0;
    while(used[ty]) ++ty;
    rep(i,8){
        bool flag=true;
        rep(j,cur){
            if(i==x[j]) flag=false;
            if(abs(ty-y[j])==abs(i-x[j])) flag=false;
        }
        if(flag){
            x[cur]=i,y[cur]=ty;
            used[ty]=true;
            dfs(cur+1);
            used[ty]=false;
        }
    }
}
string s[8];
int main(){
    int k;
    cin>>k;
    rep(i,k){
        cin>>y[i]>>x[i];
        used[y[i]]=true;
    }
    dfs(k);
    rep(i,8)rep(j,8) s[i]+='.';
    rep(i,8) s[ay[i]][ax[i]]='Q';
    rep(i,8) cout<<s[i]<<endl;
    return 0;
}
