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>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef pair<ll, ll> pll;
typedef pair<int, int> pii;
typedef double db;
#define all(x) (x).begin(), (x).end()

const ll MAXN = 1e9, MINN = -1e9;
const ld pi = 3.14159265358979323846264338327950288;
const ll lcm_16 = 720720;

void fastio() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
}

ll gcd(ll a, ll b) {
    while (b) {
        a %= b;
        swap(a, b);
    }
    return a;
}

void solve() {
    int n, m;
    cin >> n >> m;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            ll x;
            cin >> x;
            if ((i + j) % 2) {
                cout << lcm_16 << " ";
            } else {
                cout << lcm_16 + x * x * x * x << " ";
            }
        }
        cout << "\n";
    }
}

int main() {
    ll t = 1;
    fastio();
    // cin >> t;
    while (t--) { solve(); }
    return 0;
}