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 double ld;
const long long inf = 1100000000000000000;
#define maxn 1000005
#define mod 1000000007 //998244353
#define int long long
#define PB push_back
#define eb emplace_back
#define all(c) (c).begin(),(c).end()
#define pii pair <int,int>
#define vi vector<int>
#define F first
#define S second
#define rep(i,a,b)  for(int i=a;i<b;i++)
#define rep1(i,a,b) for(int i=a;i<=b;i++)
#define repd(i,a,b) for(int i=a;i>=b;--i)
#define sz(x) (int)((x).size())  
#define mem(a,x) memset(a,x,sizeof(a))
const long double PI=3.141592653589793238462643383279502884197169399375105820974944;

void solve(){
	int n, m; cin>>n>>m;
	int z = 720720; //lcm{1..16}
	int a[n][m];
	rep(i, 0, n)	
		rep(j, 0, m)	
			cin>>a[i][j];
	rep(i, 0, n){
		rep(j, 0, m){
			int k = z-pow(a[i][j], 4);
			if((i+j)%2)
				cout<<z<<" ";
			else
				cout<<k<<" ";
		}cout<<"\n";
	}
}

signed main(){
    ios_base::sync_with_stdio(false), cin.tie(nullptr); 
    int T=1;// cin>>T; 
    while(T--){
        solve();cout<<'\n';
    }
    return 0;
}