Problem: You are given a matrix a consisting of positive integers. It has n rows and m columns.

Construct a matrix b consisting of positive integers. It should have the same size as a, and the following conditions should be met: 

  * 1 ≤ b_{i,j} ≤ 10^6; 
  * b_{i,j} is a multiple of a_{i,j}; 
  * the absolute value of the difference between numbers in any adjacent pair of cells (two cells that share the same side) in b is equal to k^4 for some integer k ≥ 1 (k is not necessarily the same for all pairs, it is own for each pair). 



We can show that the answer always exists.

Input

The first line contains two integers n and m (2 ≤ n,m ≤ 500).

Each of the following n lines contains m integers. The j-th integer in the i-th line is a_{i,j} (1 ≤ a_{i,j} ≤ 16).

Output

The output should contain n lines each containing m integers. The j-th integer in the i-th line should be b_{i,j}.

Examples

Input


2 2
1 2
2 3


Output


1 2
2 3


Input


2 3
16 16 16
16 16 16


Output


16 32 48
32 48 64


Input


2 2
3 11
12 8


Output


327 583
408 664

Note

In the first example, the matrix a can be used as the matrix b, because the absolute value of the difference between numbers in any adjacent pair of cells is 1 = 1^4.

In the third example: 

  * 327 is a multiple of 3, 583 is a multiple of 11, 408 is a multiple of 12, 664 is a multiple of 8; 
  * |408 - 327| = 3^4, |583 - 327| = 4^4, |664 - 408| = 4^4, |664 - 583| = 3^4. 

Solution (in CPP):

Solution:
#include <iostream>
#include<map>
#include<vector>
#include<climits>
#include<queue>
#include<cmath>
#include<stack>
#include<string>
#include<algorithm>
#define pb push_back
#define N (lli)(3e5)
#define lli  long long int
#define mod (lli)(998244353)
#define INF (1e9)
#define fast_io ios_base::sync_with_stdio(false);cin.tie(NULL)
#define pb push_back
using namespace std;
// Jai Hanuman Gyan Gun Sagar Jai Kapis Ti Log Ujagar
int main()
{
    int n,m;
    cin>>n>>m;
    vector<vector<int>>arr(n,vector<int>(m));
    vector<vector<int>>soln(n,vector<int>(m));
    for(int i=0;i<n;i++)
        for(int j=0;j<m;j++)
        {
            cin>>arr[i][j];
            if((i+j)%2==0) soln[i][j] = 720720;
            else soln[i][j] = 720720 + arr[i][j]*arr[i][j]*arr[i][j]*arr[i][j];
        }
    
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<m;j++)  cout<<soln[i][j]<<" ";
        cout<<endl;
    }
    
    
    
    
    
}