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 <stdio.h>
int a[501][501];
int b[501][501];
int gcd(int a, int b) {//a<b
	if (a > b) a ^= b ^= a ^= b;
	if (a == 0) return b;
	return gcd(b%a, a);
}
int main() {
	int i, j, n, m;
	int tgt[17] = { 0 };
	for (scanf("%d%d", &n, &m), i = 1; i <= n; ++i)
		for (j = 1; j <= m; ++j) scanf("%d",a[i]+j);
	int lcm = 2;
	for (i = 3; i <= 16; ++i) lcm = lcm * i / gcd(lcm,i);
	tgt[1] = lcm - 1;
	for (int i = 1; i <= 32; ++i) {
		int d = i * i*i*i;
		int x = lcm - d;
		if (x > 0) {
			for (j = 2; j <= 16; ++j) if (x%j == 0) tgt[j] = x;
		}
		x = lcm + d;
		if (x <= 1000000) {
			for (j = 2; j <= 16; ++j) if (x%j == 0) tgt[j] = x;
		}
	}
	for (i = 1; i <= n; ++i) {
		for (j = 1; j <= m; ++j) {
			if ((i + j) & 1) printf("%d ",lcm);
			else printf("%d ",tgt[a[i][j]]);
		}
		puts("");
	}

	return 0;
}