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>
#define ls(p) p<<1
#define rs(p) p<<1|1
#define de(x) cout<<#x<<" = "<<x<<endl;
#define rep(i,a,b) for(int i = (a);i<=(b);++i)
#define endl '\n'
#define PI acos(-1.0)
#define pb push_back
using namespace std;
void debug(){freopen("data.in","r",stdin);freopen("data.out","w",stdout);}
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
const int maxn  = 1e6+10;
const int INF = 1e6;
const int val = 720720;
int n,m;
int a[501][501];

int main(){
	scanf("%d%d",&n,&m);
	rep(i,1,n)rep(j,1,m){
		int v; scanf("%d",&v);
		if((i + j)&1) a[i][j] = val + v*v*v*v;
		else a[i][j] = val;
	} 
	rep(i,1,n)rep(j,1,m)
		printf("%d%c",a[i][j],j==m?'\n':' ');
	return 0;
}
//720720