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 fix(f,n) std::fixed<<std::setprecision(n)<<f
#define fast ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
typedef long long int ll;
typedef unsigned long long int ull;
#define vi vector<int>
#define pii pair<int,int>
#define vii vector<pii>
#define max(a,b) ((a>b)?a:b)
#define min(a,b) ((a>b)?b:a)
#define max3(a,b,c) ((a>b)?(a>c)?a:c:(b>c)?b:c)
#define min3(a,b,c) ((a<b)?(a<c)?a:c:(b<c)?b:c)
#define REP(i,a,n) for(ll i=a;i<n;i++)
#define pb push_back
#define mp make_pair
#define mod 1000000007
#define MAX 1000000001

ll arr[200001],dp[200001]={},par[200001];

void dfs(int u,int parent,vector<vector<int>> &g,int lvl,vector<vector<int>> &nodes){
    par[u]=parent;
    nodes[lvl].pb(u);
    for(auto i:g[u]){
        if(i==parent){
            continue;
        }
        dfs(i,u,g,lvl+1,nodes);
    }
}

int main(){
	fast;
	int test=1;
	//cin >> test;
	while(test--){
	    int n,m;
	    cin >> n >> m;
	    for(int i=0 ; i<n ; i++){
	        for(int j=0 ; j<m ; j++){
	            int x;
	            cin >> x;
	            if((i%2)==(j%2)){
	                cout << 720720+(x*x*x*x) << " ";
	            }else{
	                cout << 720720 << " ";
	            }
	        }
	        cout << endl;
	    }
	}
	return 0;
}