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>
using namespace std;
const int N = 8;
int st[N][N]={};
int rc[N]={},cc[N]={};
int sc[N*2]={},dc[N*2]={};
int cnt=0;
void print(){
  for(int i=0;i<N;i++){
    for(int j=0;j<N;j++)
      cout<<(st[i][j]?"Q":".");
    cout<<endl;
  }
  exit(0);
}

void dfs(int y,int x){
  if(y==N){
    if(cnt==N) print();
    return;
  }
  int ny=y+(x+1)/N,nx=(x+1)%N;
  if(st[y][x]==2){
    dfs(ny,nx);
    return;
  }
  
  dfs(ny,nx);
  if(!rc[y]&&!cc[x]&&!sc[y+x]&&!dc[N+y-x]){
    rc[y]++;
    cc[x]++;
    sc[y+x]++;
    dc[N+y-x]++;

    st[y][x]=1;
    cnt++;
    dfs(ny,nx);
    cnt--;
    st[y][x]=0;
    
    rc[y]--;
    cc[x]--;
    sc[y+x]--;
    dc[N+y-x]--;
  }  
}

int main(){
  int n;
  cin>>n;
  for(int i=0;i<n;i++){
    int r,c;
    cin>>r>>c;
    st[r][c]=2;
    rc[r]++;
    cc[c]++;
    sc[r+c]++;
    dc[N+r-c]++;
    cnt++;
  }  
  dfs(0,0);
  return 0;
}

