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;
#define pb push_back
#define sz size()
#define ff first
#define ss second
#define mod 1000000007
#define pie 3.14159265359
#define minheap priority_queue<int , vector<int> , greater<int> > 
#define maxheap priority_queue<int> 	
#define fast ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL) ;
ll gcd(ll a, ll b)
{
	if(b==0)
	return a;
	else return gcd(b,a%b);
}
int main()
{
	fast;
	ll t;
	t=1;
	while(t--)
	{
		ll n,m,i,j;
		cin>>n>>m;
		ll g=1;
		for(i=1;i<17;i++)
		{
			g=g*i/gcd(g,i);
		}
		for(i=0;i<n;i++)
		{
			for(j=0;j<m;j++)
			{
				ll x;
				cin>>x;
				if((i+j)%2==0)
				{
					cout<<g<<" ";
				}
				else
				{
					cout<<g+x*x*x*x<<' ';
				}
			}
			cout<<"\n";
		}
		
    }
}