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>
#include<iostream>
using namespace std;
#define fastio ios_base::sync_with_stdio(0);cin.tie(0)
#define FOR(i,m,n) for(int i = (m); i < (n); i++)
#define pb push_back
#define mp make_pair
#define fst first
#define snd second
#define all(v) sort(v.begin(),v.end())
#define sz(v) int(v.size())
const int MOD=1e9+7;
const int mod= 998244353;
typedef long long ll;
typedef vector<int> vi; 
typedef vector<ll> vll; 
typedef pair<ll,ll> ii;
int main() {
		int n,m;
		cin>>n>>m;
		int a[700][700];
		FOR(i,0,n){
			FOR(j,0,m){
				cin>>a[i][j];
			}
		}
		int b[700][700];
		FOR(i,0,n){
			FOR(j,0,m){
				if((i+j)%2){
					b[i][j]=720720+a[i][j]*a[i][j]*a[i][j]*a[i][j];
				}
				else b[i][j]=720720;
			}
		}
		FOR(i,0,n){
			FOR(j,0,m){
				cout<<b[i][j]<<" ";
			}
			cout<<endl;
		}
	return 0;
}