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;
const int maxn=3e5+9;
const int MAX=6e5+9;
const int MAX1=5e6+9;
const int INF=0x3f3f3f3f;
const long double ep=1e-12;
const double ep1=1e-14;
const int mod=998244353;
const double inf=1e20;
const long double pi=acos(-1);
#define debug1 puts("?");
#define debug(x) cout<<"##"<<(x)<<endl;
#define mk make_pair
#define PII pair<int,int>
#define PDI pair<double,int>
#define PIII pair<int,PII >
#define PIII1 pair<PII,int>
#define PIIII pair<PII,PII >
#define PIL pair<int,ll>
#define PLI pair<ll,int>
#define PLIII pair<PLI,PII >
#define PLL pair<ll,ll>
#define fi first
#define se second
#define pb push_back
#define eb emplace_back
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define dep(i,a,b) for(int i=(a);i>=(b);--i)
#define sd(x) scanf("%d",&(x))
#define slld(x) scanf("%lld",&(x))
#define sdd(x,y) scanf("%d%d",&(x),&(y))
#define sc(s) scanf("%s",(s))
#define pd(x) printf("%d\n",(x))
#define plld(x) printf("%lld\n",(x))
#define pdk(x) printf("%d ",(x))
#define plldk(x) printf("%lld ",(x))
#define pdd(x,y) printf("%d %d\n",(x),(y))
#define PR(x) printf("Case %d: ",(x))
#define INF 0x3f3f3f3f
typedef long double ld;
typedef unsigned long long ull;
/*
给你一个n*m的矩阵a
需要构造一个n*m的矩阵b
bij是aij的倍数
1<=bij<=1e6
对于某一个位置，它跟相邻4个数的差值的绝对值都为k^4(爬
对于任意一对相邻的两个数，差值都为某个正数的4次方 
1-16的最小公倍数为720720 
*/
int f[20],n,m,a;
void solve()
{
	rep(i,1,16)
	{
		rep(k,1,30)
		{
			if((720720-k*k*k*k)%i==0)
			{
				f[i]=720720-k*k*k*k;
				break;
			}
		}
	}
	sdd(n,m);
	rep(i,1,n)
	{
		rep(j,1,m)
		{
			sd(a);
			if((i+j)&1)pdk(720720);
			else pdk(f[a]);
		}
		puts("");
	}
}
int main()
{
//    ios::sync_with_stdio(false);
//    cin.tie(0);
    int T=1;
//    sd(T);
    while(T--)solve(); 
}