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 <vector>
#include <string>
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#define LL long long
using namespace std;
inline LL rin()
{
    LL s=0;
    bool bj=false;
    char c=getchar();
    for(;(c>'9'||c<'0')&&c!='-';c=getchar());
    if(c=='-')bj=true,c=getchar();
    for(;c>='0'&&c<='9';c=getchar())s=(s<<1)+(s<<3)+(c^'0');
    if(bj)return -s;
    return s;
}
inline int min(int x,int y){return x<y?x:y;}
inline int max(int x,int y){return x>y?x:y;}
inline LL min(LL x,LL y){return x<y?x:y;}
inline LL max(LL x,LL y){return x>y?x:y;}
inline void jh(int &x,int &y){if(x!=y)x^=y^=x^=y;return;}
inline void jh(LL &x,LL &y){if(x!=y)x^=y^=x^=y;return;}

inline int Gcd(int x,int y){return (!y)?(x):(Gcd(y,x%y));}
inline int Lcm(int x,int y){return x/Gcd(x,y)*y;}

const int N=5e2+3;
int n,m;
int L=1;

int a[N][N];

inline void work()
{
    n=rin();m=rin();
    for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)a[i][j]=rin();
    for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){int x=a[i][j];if((i+j)&1)printf("%d ",L);else printf("%d ",L+x*x*x*x);}printf("\n");}
    return;
}
int main()
{
    for(int i=1;i<=16;i++)L=Lcm(L,i);
    work();
    return 0;
}