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 <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <cmath>
#include <algorithm>
#define ll long long
#define rep(i, a, b) for (int i = a; i <= (int)b; i ++)
#define per(i, a, b) for (int i = (int)a; i >= b; i --)
using namespace std;
int n, m;
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int main()
{
    int tmp = 2;
    rep(i, 3, 16) tmp = i / gcd(tmp, i) * tmp;
    scanf("%d %d", &n, &m);
    rep(i, 1, n) rep(j, 1, m) {
        int x; scanf("%d", &x);
        if((i + j) % 2) x = tmp + x * x * x * x;
        else x = tmp;
        printf("%d%c", x, " \n"[j == m]);
    }
    return 0;
}