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:
// how to come up with these logics :(

#include <bits/stdc++.h>
    using namespace std;
     
    #define fast { ios :: sync_with_stdio(false); cin.tie(0); cout.tie(0); }
    #define pb push_back
    #define ll long long
    #define ld long double 
    #define vi vector<int>
    #define vll vector<long long>
    #define mp make_pair 
    #define infi INT_MAX
    #define infl LONG_LONG_MAX
    #define f(i,a,b) for(ll i=a;i<b;i++)
    #define ff first
    #define ss second
    #define pll pair<ll,ll> 
    #define endl "\n"
    #define ALL(v) v.begin(),v.end()
    #define nl cout<<"\n";
   
 
 
     int main() 
    {
        fast
       
        
     
        ll ntc=1;
     //  cin>>ntc;
     
        f(ks,1,ntc+1)
        {
          //cout<<"Case #"<<i<<": ";
          ll n,m; cin>>n>>m;
          ll a[505][505];
          f(i,0,n){
          	f(j,0,m){
          		cin>>a[i][j];
          		ll x = a[i][j];
          		if((i+j)%2==0) cout<<"720720"<<" ";
          		else cout<<x*x*x*x + 720720<<" ";
          	}
          	cout<<endl;
          }
          
         
          
        }
     
        return 0;
    }