// Author: Klay Thompson
// Problem: P10004 [集训队互测 2023] Permutation Counting 2
// Memory Limit: 512 MB
// Time Limit: 5000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>

#define rep(i, j, k) for (int i = (j); i <= (k); ++i)
#define per(i, j, k) for (int i = (j); i >= (k); --i)
#define SZ(v) int((v).size())
#define ALL(v) (v).begin(),(v).end()
#define fi first
#define se second
using ll = long long;
using pii = std::pair<int, int>;
using pll = std::pair<ll, ll>;

template<class T>inline void chkmn(T &x, T y) { if (y < x) x = y; }
template<class T>inline void chkmx(T &x, T y) { if (y > x) x = y; }

using namespace std;

const int maxn = 510;
const int maxm = 500010;

int n, mod, f[maxn][maxn], g[maxn][maxn], h[maxn], fac[maxm], ivf[maxm];

__int128 mu;
namespace barrett {
  inline ll reduce(ll x) {
    ll r = x - (mu * x >> 64) * mod;
    return r >= mod ? r - mod : r;
  }
  inline void setmod(ll mod) {
    mu = -1ull / mod;
  }
}
using namespace barrett;

int binom(int x, int y) {
  if (x < 0 || y < 0 || x < y) return 0;
  return reduce(1ll * reduce(1ll * fac[x] * ivf[y]) * ivf[x - y]);
}

int coe(int x) {
  return (x & 1) ? mod - 1 : 1;
}

inline void add(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}

inline int fpw(int x, int p) {
  int res = 1;
  while (p) {
    if (p & 1) res = 1ll * res * x % mod;
    x = 1ll * x * x % mod, p >>= 1;
  }
  return res;
}

int main() {
  cin.tie(nullptr) -> ios::sync_with_stdio(false);
  cin >> n >> mod, setmod(mod);
  fac[0] = 1;
  rep (i, 1, maxm - 1) fac[i] = 1ll * fac[i - 1] * i % mod;
  ivf[maxm - 1] = fpw(fac[maxm - 1], mod - 2);
  per (i, maxm - 1, 1) ivf[i - 1] = 1ll * ivf[i] * i % mod;
  rep (p, 1, n) {
    rep (j, 1, n) h[j] = 0;
    rep (j, 1, n) rep (q, 1, j) add(h[j], reduce(1ll * reduce(1ll * coe(j - q) * binom(j, q)) * binom(n + p * q - 1, n)));
    rep (i, p, n) rep (j, 1, n) add(f[i][j], reduce(1ll * reduce(1ll * coe(i - p) * binom(i, p)) * h[j]));
  }
  rep (p, 1, n) {
    rep (j, 1, n) h[j] = 0;
    rep (j, 1, n) rep (q, 1, j) add(h[j], reduce(1ll * reduce(1ll * coe(j - q) * f[p][q]) * binom(n - q, j - q)));
    rep (i, p, n) rep (j, 1, n) add(g[i][j], reduce(1ll * reduce(1ll * coe(i - p) * binom(n - p, i - p)) * h[j]));
  }
  per (i, n, 1) per (j, n, 1) cout << g[i][j] << " \n"[j == 1];
}