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>

#pragma GCC optimize("Ofast")
#pragma GCC optimize("O4")
#pragma GCC optimize("no-stack-protector")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("unswitch-loops")
#pragma GCC optimize("fast-math")
#pragma GCC optimize("tracer")
#pragma GCC target("sse2")
#pragma GCC target("sse3")
#pragma GCC target("popcnt")
#pragma GCC target("abm")
#pragma GCC target("mmx")
#pragma GCC target("tune=native")

namespace Competitive {
using namespace std;
using ll = long long;
using ld = long double;
using ull = unsigned long long;

#define all(a) begin(a), end(a)
#define len(a) (int)a.size()
#define pb push_back
#define eb emplace_back
#define em emplace

void useiostream() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cout.setf(ios::fixed), cout.precision(20);
}

void drop(char const *bu) {
  printf("%s", bu);
  exit(0);
}

ull gcd(ull a, ull b) { return b == 0 ? a : gcd(b, a % b); }

ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }

ld const pi = acos(-1.0L);
ld const eps = 1e-6;

template <typename T, typename TT>
bool mi(T &a, TT b) {
  return a > b ? (a = b, true) : false;
}
template <typename T, typename TT>
bool ma(T &a, TT b) {
  return a < b ? (a = b, true) : false;
}

auto seed = chrono::high_resolution_clock::now().time_since_epoch().count();
mt19937 ran(seed);

int Mod[] = {1000000007, 1052339657, 998244353};
int const mod = 1000000007;

}  // namespace Competitive

using namespace Competitive;

int n, m;
int arr[505][505];

ll lcm(vector<ll> a) {
  if (len(a) == 2) return lcm(a[0], a[1]);
  auto t = a.back();
  a.pop_back();
  return lcm(t, lcm(a));
}

int main(int argc, char *args[]) {
  // freopen("input.txt", "r", stdin);
  // freopen("output.txt", "w", stdout);
  useiostream();
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> arr[i][j];
    }
  }
  vector<ll> a;
  for (int i = 1; i <= 16; i++) a.pb(i);
  ll t = lcm(a);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (i + j & 1) {
        cout << t + arr[i][j] * arr[i][j] * arr[i][j] * arr[i][j];
      } else {
        cout << t;
      }
      cout << " ";
    }
    cout << "\n";
  }
  return 0;
}