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:
//#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>

using namespace std;

//vector string deque break continue
#define forn(i, s, f) for (int i = (int)s; i < (int)f; i++)
#define ll long long
#define ull unsigned long long
#define ld long double
#define pii pair <int, int>
#define fs first
#define sc second
#define pf push_front
#define pb push_back
#define pop_f pop_front
#define pop_b pop_back
#define eb emplace_back
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define sz(x) (int)(x).size()

#ifdef DEBUG
#else
    #define cerr if (false) cerr
#endif

template <typename T> istream& operator>>(istream& in, vector <T>& a) {for (auto& i : a) in >> i; return in;}
template <typename T> ostream& operator<<(ostream& out, vector <T>& a)  {for (auto& i : a) out << i << " "; return out;}
template <typename T, typename U> void chkmin(T& a, U b) {if (a > b) a = b;}
template <typename T, typename U> void chkmax(T& a, U b) {if (a < b) a = b;}

constexpr int LCM = 720720;

int main() {
    ios_base::sync_with_stdio(0), cin.tie(0);
    int n, m;
    cin >> n >> m;
    vector<vector<int>> a(n, vector<int> (m));
    cin >> a;
    forn (i, 0, n) {
        forn (j, 0, m) {
            if (i + j & 1)
                cout << LCM;
            else
                cout << LCM + a[i][j] * a[i][j] * a[i][j] * a[i][j];
            cout << " ";
        }
        cout << "\n";
    }
    return 0;
}