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 <vector>
#include <chrono>
#include <random>
#include <cassert>

std::mt19937 rng((int) std::chrono::steady_clock::now().time_since_epoch().count());

long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }

int main() {
	std::ios_base::sync_with_stdio(false); std::cin.tie(NULL);
	long long L = 1;
	for(int i = 1; i <= 16; i++) {
		L = lcm(i, L);
	}
	std::vector<long long> val(17, 0);
	for(int i = 1; i <= 16; i++) {
		int j = 0;
		int k = 1;
		while(1) {
			j += i;
			while(k * k * k * k < j) k++;
			if(k * k * k * k == j) break;
		}
		val[i] = L + j;
		//std::cout << val[i] << '\n';
	}
	int n, m;
	std::cin >> n >> m;
	for(int i = 0; i < n; i++) for(int j = 0; j < m; j++) {
		int x;
		std::cin >> x;
		std::cout << ((i + j) % 2 == 0 ? L : val[x]) << (j + 1 == m ? '\n' : ' ');
	}
}