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:
// Igorjan94, template version from 13 October 2017. C++17 version, modified 18 march 2020 (writeln<tuple>, whole->all) {{{
#include <bits/stdc++.h>

using namespace std;

#define FOR(i, m, n) for (int i = m; i <  (int) (n); ++i)
#define ROF(i, m, n) for (int i = m; i >= (int) (n); --i)
#define forn(i, n)   for (int i = 0; i < (int) (n); ++i)
#define  fori1(n)    for (int i = 1; i < (int) (n); ++i)
#define  forj1(n)    for (int j = 1; j < (int) (n); ++j)
#define   fori(n)    for (int i = 0; i < (int) (n); ++i)
#define   forj(n)    for (int j = 0; j < (int) (n); ++j)
#define     SZ(a)    int(size(a))

typedef  pair<int, int>   pii;
typedef   vector<int>     vi;
typedef    long long      ll;

#define pb push_back
#define all(a) begin(a), end(a)
#define ints(a...) int a; readln(a)

[[maybe_unused]] const int MOD = 1000000007;
[[maybe_unused]] const int INTMAX = numeric_limits<int>::max();

#define ttt12i template<class T1, class T2> inline
#define  ttti  template<class T> inline

void writeln(){cout<<"\n";}ttti void print(T&& a);ttti void priws(T&& a);ttti void read(T& a);
template<class... Args> inline void readln(Args&... args){(read(args),...);}
template<class H, class...T> inline void writeln(H&& h,T&&...t){priws(h);(print(t),...);writeln();}

//Igorjan
//}}}

void run()
{
    ints(n, m);
    vector<vi> a(n, vi(m));
    vector<vi> b(n, vi(m, 720720));
    readln(a);
    fori(n)
        forj(m)
            if ((i + j) % 2 == 0)
                b[i][j] += a[i][j] * a[i][j] * a[i][j] * a[i][j];
    writeln(b);
}

//{{{
int main()
{
    ios_base::sync_with_stdio(false);
    run();
    cerr << fixed << setprecision(0) << "Execution time = " << 1000.0 * clock() / CLOCKS_PER_SEC << "ms\n";
    return 0;
}

#define a _a
#define n _n
ttt12i ostream&operator<<(ostream&os,pair<T1,T2>const&a);
template<typename T,typename D=decltype(*begin(declval<T>())),typename enable_if<!is_same<T,basic_string<char>>::value>::type* =nullptr>
ostream&operator<<(ostream&os,T const&a){auto it=begin(a);if(it!=end(a))os<<*it++;while(it!=end(a))os<<"\n "[is_fundamental<typename T::value_type>::value]<<*it++;return os;}
ttt12i ostream&operator<<(ostream&os,pair<T1,T2>const&a){return os<<a.first<<" "<<a.second;}
ttt12i istream&operator>>(istream&is,pair<T1,T2>&a){return is>>a.first>>a.second;}
ttti   istream&operator>>(istream&is,vector<T>&a){fori(a.size())is>>a[i];return is;}
ttti void print(T&&a){cout<<" "<<a;}
ttti void priws(T&&a){cout<<a;}
ttti void read(T&a){cin>>a;} //}}}