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>
#include <unordered_map>
#include <chrono>

#define mp make_pair
#define ll long long
#define ll_MAX 1e18

using namespace std;

vector<int> nums;
double quadraticFormula(double a, double b, double c)
{
    pair<double, double> totalResults = mp(-1, -1);
    double numeratorSqrtResult = sqrt((b * b) - (4 * a * c));
    double denominatorResult = 2 * a;
    double addNegativeTo_b = 0 - b;

    if (denominatorResult == 0)
    {
        cout << 1 / 1;
        exit(0);
    }

    totalResults.first = (addNegativeTo_b + numeratorSqrtResult) / denominatorResult;
    totalResults.second = (addNegativeTo_b - numeratorSqrtResult) / denominatorResult;
    return max(totalResults.first, totalResults.second);
}
int main() {

    ios::sync_with_stdio(false);
    cin.tie(NULL);

    int n, m; cin >> n >> m;

    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            int x; cin >> x;
            if ((i+j)%2==0) {
                cout << 720720 << " ";
            }
            else {
                cout << 720720 - pow(x, 4) << " ";
            }
        }
        cout << endl;
    }

    return 0;
}