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 ll long long
#define ld long double
#define For(i,x,y)	for(ll i=(x);i<=(y);++i)
#define FOr(i,x,y)	for(ll i=(x);i>=(y);--i)
#define rep(i,x,y)	for(ll i=(x);i<(y);++i)
#define clr(a,v)	memset(a,v,sizeof(a))
#define cpy(a,b)	memcpy(a,b,sizeof(a))
#define fi	first
#define se	second
#define	pb	push_back
#define mk	make_pair
#define pa	pair<ll,ll>
#define y1	y11111111111111
#define debug	puts("@@@@@@@@@@@@@@@@@@@@@@@")
ll read(){
	ll x=0,f=1;	char ch=getchar();	
	for(;ch<'0'||ch>'9';ch=getchar())	if (ch=='-')	f=-1;
	for(;ch>='0'&&ch<='9';ch=getchar())	x=x*10+ch-'0';
	return x*f;
}
void write(ll x){
	if (x<0)	putchar('-'),write(-x);
	else{
		if (x>=10)	write(x/10);
		putchar(x%10+'0');
	}
}
const ll N=300300,mod=1e9+7;
ll n,m;
int main(){
	n=read();	m=read();
	ll lcm=360360*2;
	For(i,1,n){
		For(j,1,m){
			ll x=read();
			write(((i+j)&1)*x*x*x*x+lcm),putchar(' ');
		}
		puts("");
	}
}