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;
using ll = long long;
using ld = long double;
using uint = unsigned int;
using ull = unsigned long long;
template <typename T>
using pair2 = pair<T, T>;
using pii = pair<int, int>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
using mii = map<int, int>;
using mll = map<ll, ll>;
using vi = vector<int>;
using vl = vector<ll>;
struct Comp {
  bool operator()(const std::pair<int, int> &a, const std::pair<int, int> &b) {
    if (a.first != b.first) {
      return a.first > b.first;
    }
    return a.second < b.second;
  }
};
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
ll myRand(ll B) { return (ull)rng() % B; }
bool sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second < b.second);
}
clock_t startTime;
double getCurrentTime() {
  return (double)(clock() - startTime) / CLOCKS_PER_SEC;
}
void solve() {
  int n, s;
  cin >> n >> s;
  deque<int> v;
  for (int i = 0; i < n - 1; i++) {
    v.push_back(0);
  }
  v.push_back(s);
  for (int i = 0; i < n; i++) {
    for (auto j : v) {
      cout << j << " ";
    }
    int l = v.back();
    v.pop_back();
    v.push_front(l);
    cout << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL), cout.tie(NULL);
  solve();
}