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=5e2+7;
const int modn = 1e9+7;
int n,m;
int a[maxn][maxn];
int four[maxn];
int gongbei;
int gcd(int a,int b){
    if(b == 0)return a;
    return gcd(b,a%b);
}
void init(){
    for(int i = 1;i < maxn;i++){
        ll res = i*i*i*i;
        if(res > 1000000)break;
        four[i] = res;
//        printf("four[%d]: %d\n",i,four[i]);
    }
    gongbei = 1;
    for(int i = 2;i <= 16;i++){
        gongbei = gongbei * i / gcd(gongbei,i);
    }
//    printf("%d\n",gongbei);
}
int main(){
    scanf("%d%d",&n,&m);
    for(int i = 1;i <= n;i++){
        for(int j = 1;j <= m;j++){
            scanf("%d",&a[i][j]);
        }
    }
    init();
    for(int i = 1;i <= n;i++){
        for(int j = 1;j <= m;j++){
            if((i+j)&1)a[i][j] = gongbei - four[a[i][j]];
            else a[i][j] = gongbei;
//            printf("%d %d\n",gongbei,a[1][2]);
//            printf("a[%d][%d]: %d\n",i,j,a[i][j]);
        }
    }
    for(int i = 1;i <= n;i++){
        for(int j = 1;j < m;j++){
            printf("%d ",a[i][j]);
        }
        printf("%d\n",a[i][m]);
    }
}