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>
//#include <ext/pb_ds/assoc_container.hpp>
//using namespace __gnu_pbds;
using namespace std;
#define int      long long int
#define MOD      998244353
#define pb       push_back
#define pf       push_front
#define vi       vector<int>
#define mi       map<int,int>
#define umi      unordered_map<int,int>
#define pii      pair<int,int>
#define ff       first
#define ss       second
#define inf      1e18
#define case    int test;cin>>test;while(test--)
//typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> pbds;
int powm(int x, int y, int m = MOD) {
	x = x % m; int res = 1; while (y)
	{if (y & 1)res = res * x; res %= m; y = y >> 1; x = x * x; x %= m;} return res;
}
int modi(int a, int m = MOD) {return powm(a, m - 2, m);}
void pre()
{
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
}
int32_t main()
{	pre();
//case
	{	int n, m;
		cin >> n >> m;
		int a[n][m];
		for (int i = 0; i < n; i++)
			for (int j = 0; j < m; j++)
				cin >> a[i][j];
		int vl = 720720;
		for (int i = 0; i < n; i++)
		{
			for (int j = 0; j < m; j++)
			{
				if ((i + j) % 2 == 0)
					cout << vl << " ";
				else
				{
					cout << vl + pow(a[i][j], 4) << " ";
				}
			}
			cout << endl;
		}
	}
	return 0;
}