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;
#define ll long long
#define ld long double
#define pb push_back
#define forq(i , a , b) for (int i = (a); i <= (b); ++i)
#define qrof(i , b , a) for (int i = (a); i >= (b); --i)
#define forr(i , b) forq( i , 0 , b - 1 )
#define F first
#define S second
#define IF ->first
#define IS ->second
#define endl '\n'
#define qqmemset(array , val) memset (array , val , sizeof(array))
#define ALLV(vect) vect.begin() , vect.end()
#define mid (st + en) / 2
#define mid1 (2 * st + en) / 3
#define mid2 (2 * en + st) / 3
#define lef 2 * Node
#define rig lef + 1
mt19937 rng( chrono::steady_clock::now().time_since_epoch().count() );
#define Ran(a, b) rng() % ( (b) - (a) + 1 ) + (a)

ll R = 7 + 1e9 , R1 = 19491001 , R2 = 236 , NUMTESTCASE ;
const ll NN = 10 + 1e6 ;
const double pi = acos(-1.0) ;
int di [8] = {1 , 0 , -1 , 0  , 1 , -1 , 1  , -1 } , dj [8] = {0 , 1 , 0  , -1 , 1 , -1 , -1 , 1  } ;

int main()
{
    ios_base::sync_with_stdio(false); cin.tie(NULL);cout.tie(NULL) ;
    int n , m , x ;
    cin >> n >> m ;
    forq (i , 1 , n) {
        forq (j , 1 , m) {
            cin >> x ;
            cout << 720720 + ((i + j) % 2) * x * x * x * x  << " " ;
        }
        cout << endl ;
    }
    return 0;
}