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>
#define M 1000000007
#define MOD 1003
#define MD 1337
#define MAXN 200005
#define F first
#define S second
#define int int64_t
#define filler(arr, n)          \
    for (int i = 0; i < n; i++) \
        cin >> arr[i];
using namespace std;

vector<vector<int>> directions { { 1, 0 }, { -1, 0 }, { 0, 1 }, { 0, -1 } };
void solve();
int32_t main()
{
    int T = 1;
    // cin >> T;

    while (T--) {
        solve();
    }
    return 0;
}
/*
pre processing
binary search
target sum
dp
pattern
*/

void solve()
{
    int n, m;
    cin >> n >> m;
    vector<vector<int>> arr(n, vector<int>(m, 0)), ans(n, vector<int>(m, 0));
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            cin >> arr[i][j];
        }
    }
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            if ((i + j) % 2 == 0) {
                cout << 720720 << " ";
            } else {
                cout << 720720 + (int)pow(arr[i][j], 4) << " ";
            }
        }
        cout << endl;
    }
}