#include <bits/stdc++.h>
#define rep(i, f, t) for(int i(f); i <= t; ++i)
#define re(i, t) for(int i(1); i <= t; ++i)
#define per(i, t, f) for(int i(t); i >= f; --i)
#define pe(i, t) for(int i(t); i >= 1; --i) 
#define nx(i, u) for(int i(head[u]); i; i = e[i].nxt)
typedef long long ll; 
using namespace std;
typedef pair <int, int> pii;
#define pb push_back
#define fi first
#define se second
#define ix(l, r) ((l + r) | (l != r)) 
#define inf (int)(1e9+7) 
bool sT;
namespace IO {
char buf[1 << 21], *p1 = buf, *p2 = buf, buf1[1 << 21];
inline char gc() {return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++;}
template<class I>
inline void read(I &x) {x = 0;I f = 1;char c = gc();while (c < '0' || c > '9') {if (c == '-') {f = -1;} c = gc();}while (c >= '0' && c <= '9') {x = x * 10 + c - '0';c = gc();}x *= f;}
template<class I>
inline void write(I x) {if (x == 0) {putchar('0');return;}I tmp = x > 0 ? x : -x;if (x < 0) {putchar('-');}int cnt = 0;while (tmp > 0) {buf1[cnt++] = tmp % 10 + '0';tmp /= 10;}while (cnt > 0)putchar(buf1[--cnt]);}
#define outn(x) write(x), putchar('\n')
#define out(x) write(x), putchar(' ')
} using namespace IO;

const int N = 505;
int n, mod;
int f[N][N];
int fac[N*N*2], inv[N*N*2];

int ksm (int a, int b) {
	ll ans = 1, base = a;
	while(b) {
		if(b & 1) ans = ans * base % mod;
		base = base * base % mod;
		b >>= 1;
	}
	return ans;
}

inline int C (int n, int m) {
	if(n < 0 || m < 0 || n < m) return 0;
	return 1ll * fac[n] * inv[m] % mod * inv[n-m] % mod;
}

int main () {
	read(n), read(mod);
	fac[0] = inv[0] = 1;
	re(i, n*n*2) fac[i] = 1ll * fac[i-1] * i % mod, inv[i] = ksm(fac[i], mod-2);
	re(i, n) re(j, n) f[i][j] = C(n + i * j - 1, i * j - 1);
	re(i, n) re(j, n) re(k, i-1) f[i][j] = (f[i][j] + mod - 1ll * f[k][j] * C(i, k) % mod) % mod;
	re(i, n) re(j, n) re(k, j-1) f[i][j] = (f[i][j] + mod - 1ll * f[i][k] * C(j, k) % mod) % mod;
	re(i, n) re(j, n) re(k, i-1) f[i][j] = (f[i][j] + mod - 1ll * f[k][j] * C(n-k, n-i) % mod) % mod;
	re(i, n) re(j, n) re(k, j-1) f[i][j] = (f[i][j] + mod - 1ll * f[i][k] * C(n-k, n-j) % mod) % mod;
	pe(i, n) {
		pe(j, n) {
			out(f[i][j]);
		}
		puts("");
	}
}