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 <bits/stdc++.h>

using namespace std;
typedef long long ll;

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    set<int> s;
    for(int i=1;i<=100;i++) s.insert(i*i*i*i);
    const int N=720720;
    vector<int> opt(17,0);
    for(int i=1;i<=16;i++)
    {
        for(int j=i;j<=N;j+=i) if(s.find(N-j)!=s.end()) opt[i]=j;
        assert(opt[i]!=0);
    }
    int n,m;
    cin >> n >> m;
    vector<vector<int>> a(n+1,vector<int>(m+1,0));
    for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) cin >> a[i][j];
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            if(((i+j)%2)==0) cout << N << " \n"[j==m];
            else cout << opt[a[i][j]] << " \n"[j==m];
        }
    }
    return 0;
}