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;
#define pb push_back
#define mp make_pair
#define FAST ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define all(x)  (x).begin(), (x).end()
typedef long long ll;
typedef long double ld;
const ll INF = 2e18;

ll mat[600][600], ans[600][600];
 
int main() {
	FAST;
	ll n, m, i, j; cin >> n >> m;
	for(i = 0; i < n; i++) {
		for(j = 0; j < m; j++) {
			cin >> mat[i][j];
		}
	}
	ll special = 720720;
	int ok = 1;
	for(i = 0; i < n; i++) {
		for(j = 0; j < m; j++) {
			if(j % 2 == ok % 2) {
				ans[i][j] = special;
			}
			else {
				ans[i][j] = special - (mat[i][j]*mat[i][j]*mat[i][j]*mat[i][j]);
			}
		}
		ok++;
	}
	for(i = 0; i < n; i++) {
		for(j = 0; j < m; j++) {
			cout << ans[i][j] << " ";
		}
		cout << "\n";
	}
	return 0;
}