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:
/*ॐ नमो भगवते वासुदेवाय नमः*/
#include<bits/stdc++.h>
using namespace std;

#define int long long
const int MOD = 1e9 + 7;
const double pi = 3.14159265359;

struct custom_hash {
    static uint64_t splitmix64(uint64_t x) {
        x += 0x9e3779b97f4a7c15;
        x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
        x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
        return x ^ (x >> 31);
    }
    size_t operator()(uint64_t x) const {
        static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
        return splitmix64(x + FIXED_RANDOM);
    }
};

int binpow(int a, int b, int m) {
    a %= m;
    int res = 1;
    while (b > 0) {
        if (b & 1)
            res = res * a % m;
        a = a * a % m;
        b >>= 1;
    }
    return res;
}
int inverse(int x){
    return binpow(x, MOD - 2, MOD);
}

void solve()
{
    int n, m, g = 720720;
    cin >> n >> m;
    vector<vector<int>> a (n, vector<int> (m));
    for(int i = 0; i < n; i++)  {
        for(int j = 0; j < m; j++)  {
            cin >> a[i][j];
            if((i + j) & 1) cout << g << " ";
            else cout << g - (a[i][j] * a[i][j] * a[i][j] * a[i][j]) << " ";
        }
        cout << "\n";
    }    
}

signed main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
    solve();
    cerr << "Time elapsed : " << 1.0 * clock() / CLOCKS_PER_SEC << " sec \n ";
}