Problem: Levko loves tables that consist of n rows and n columns very much. He especially loves beautiful tables. A table is beautiful to Levko if the sum of elements in each row and column of the table equals k.

Unfortunately, he doesn't know any such table. Your task is to help him to find at least one of them. 

Input

The single line contains two integers, n and k (1 ≤ n ≤ 100, 1 ≤ k ≤ 1000).

Output

Print any beautiful table. Levko doesn't like too big numbers, so all elements of the table mustn't exceed 1000 in their absolute value.

If there are multiple suitable tables, you are allowed to print any of them.

Examples

Input

2 4


Output

1 3
3 1


Input

4 7


Output

2 1 0 4
4 0 2 1
1 3 3 0
0 3 2 2

Note

In the first sample the sum in the first row is 1 + 3 = 4, in the second row — 3 + 1 = 4, in the first column — 1 + 3 = 4 and in the second column — 3 + 1 = 4. There are other beautiful tables for this sample.

In the second sample the sum of elements in each row and each column equals 7. Besides, there are other tables that meet the statement requirements.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long int mod = 1e9 + 7;
inline long long int add(long long int a, long long int b) {
  return (a % mod + b % mod) % mod;
}
inline long long int sub(long long int a, long long int b) {
  return (a % mod - b % mod + mod) % mod;
}
inline long long int mul(long long int a, long long int b) {
  a %= mod;
  b %= mod;
  return a * b % mod;
}
inline long long int power(long long int a, long long int b,
                           long long int MOD) {
  a %= mod;
  long long int res = 1;
  while (b) {
    if (b & 1) {
      res *= a;
      res %= MOD;
    }
    a = a * a;
    a %= MOD;
    b >>= 1;
  }
  return res;
}
clock_t time_p = clock();
void time() {
  time_p = clock() - time_p;
  cerr << "Time elapsed : " << (long double)(time_p) / CLOCKS_PER_SEC << endl;
}
signed main() {
  long long int testcase = 1;
  while (testcase--) {
    long long int n, k = 0, i, j, t, ans = 0, a[31][2];
    cin >> n >> k;
    for (i = 0; i < n; i++) {
      for (j = 0; j < n; j++) {
        if (i == j)
          cout << k << " ";
        else
          cout << 0 << " ";
      }
      cout << endl;
    }
  }
  return 0;
}