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 ft first
#define sd second
#define ll long long
#ifdef LOCAL
    #include "debughead.h"
#else
    #define dbg(...) 0
#endif

#define sz(x) (x).size()
using namespace std;


int n, m;
int a[505][505];
int b[505][505];

set <int> s1, s2;

int gcd(int a, int b) {
    if(a == 0) return b;
    return gcd(b % a, a);
}

bool vis[100];

void solve() {
    cin >> n >> m;
    for(int i = 0; i < n; i++) {
        for(int j = 0; j < m; j++) {
            cin >> a[i][j];
        }
    }

    for(int i = 0; i < n; i++) {
        for(int j = 0; j < m; j++) {
            if(i % 2 == j % 2) {
                cout << 720720 - a[i][j]*a[i][j]*a[i][j]*a[i][j] << " ";
            }
            else {
                cout << 720720 << " ";
            }
        }
        cout << "\n";
    }
}

int main()
{
    #ifndef LOCAL
        ios_base::sync_with_stdio(0); cin.tie(0);
    #endif // LOCAL
    int t = 1;
   // cin >> t;
    while(t--)
        solve();

    return 0;
}