Problem: Let us define a magic grid to be a square matrix of integers of size n × n, satisfying the following conditions. 

  * All integers from 0 to (n^2 - 1) inclusive appear in the matrix exactly once. 
  * [Bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all elements in a row or a column must be the same for each row and column. 



You are given an integer n which is a multiple of 4. Construct a magic grid of size n × n.

Input

The only line of input contains an integer n (4 ≤ n ≤ 1000). It is guaranteed that n is a multiple of 4.

Output

Print a magic grid, i.e. n lines, the i-th of which contains n space-separated integers, representing the i-th row of the grid.

If there are multiple answers, print any. We can show that an answer always exists.

Examples

Input


4


Output


8 9 1 13
3 12 7 5
0 2 4 11
6 10 15 14

Input


8


Output


19 55 11 39 32 36 4 52
51 7 35 31 12 48 28 20
43 23 59 15 0 8 16 44
3 47 27 63 24 40 60 56
34 38 6 54 17 53 9 37
14 50 30 22 49 5 33 29
2 10 18 46 41 21 57 13
26 42 62 58 1 45 25 61

Note

In the first example, XOR of each row and each column is 13.

In the second example, XOR of each row and each column is 60.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long int N = 5e5 + 5;
const long long int mod = 1e9 + 7;
const int di[4] = {-1, 0, 1, 0};
const int dj[4] = {0, -1, 0, 1};
const double pi = 3.14159265358979323846;
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
bool vis[N];
vector<long long int> adj[N];
void dfs(long long int x) {
  vis[x] = true;
  for (int i = 0; i < adj[x].size(); i++) {
    if (!vis[adj[x][i]]) {
      dfs(adj[x][i]);
    }
  }
}
long long int dis[N], ct[N];
void bfs(long long int x) {
  queue<long long int> q;
  q.push(x);
  dis[x] = 1;
  ct[x] = 1;
  while (!q.empty()) {
    long long int t = q.front();
    q.pop();
    for (auto i : adj[t]) {
      if (!dis[i]) {
        dis[i] = dis[t] + 1;
        q.push(i);
        ct[i] = 1;
      } else if (dis[i] == dis[t] + 1)
        ct[i]++;
    }
  }
}
bool cmp(pair<long long int, long long int> a,
         pair<long long int, long long int> b) {
  return a.second < b.second;
}
long long int po(long long int x, long long int n) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return po((x * x) % 998244353, n / 2);
  else
    return (x * po((x * x) % 998244353, (n - 1) / 2)) % 998244353;
}
long long int fact[N];
long long int ncr(long long int n, long long int r) {
  long long int d = fact[n];
  long long int e = (fact[n - r] * fact[r]) % 998244353;
  d *= po(e, 998244353 - 2);
  return d % 998244353;
}
bool isPrime(long long int x) {
  for (int i = 2; i <= sqrt(x); ++i)
    if (x % i == 0) return false;
  return true;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n;
  cin >> n;
  long long int a[n][n];
  long long int x = 0;
  for (int i = 0; i < n / 2; ++i) {
    for (int j = 0; j < n / 2; ++j) {
      a[i][j] = x * 4 + 1;
      a[i][j + n / 2] = x * 4 + 2;
      a[i + n / 2][j] = x * 4 + 3;
      a[i + n / 2][j + n / 2] = x * 4;
      x++;
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; ++j) cout << a[i][j] << ' ';
    cout << '\n';
  }
  return 0;
}