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 ll long long
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f3f3f
#define PI acos(-1)
#define pb push_back
#define F first
#define S second
#define lson i << 1
#define rson i << 1 | 1
#define mem(x, y) memset(x, y, sizeof(x))
#define fin(x) freopen(x, "r", stdin)
#define fout(x) freopen(x, "w", stdout)
#define fcl fclose(stdin), fclose(stdout)
#define ioss ios::sync_with_stdio(false), cout.tie(NULL)
#define pii pair<int, int>
#define vec vector
#define all(x) x.begin(), x.end()
#define lowbit(x) x &(-x)
const double eps = 1e-6;
const int MAX = 5e2 + 10;
const int MOD = 1e9 + 7;
using namespace std;

int t;
int n, m,a[MAX][MAX];
int main()
{
    ioss;
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
            cin >> a[i][j];
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= m; j++)
        {
            int k = a[i][j];
            if ((i + j) & 1)
                cout << 720720 << " ";
            else
                cout << 720720 - k * k * k * k << " ";
        }
        cout << '\n';
    }
    return 0;
}