Given is a positive integer N. Find the number of permutations (P_1,P_2,\cdots,P_{3N}) of (1,2,\cdots,3N) that can be generated through the procedure below. This number can be enormous, so print it modulo a prime number M.

* Make N sequences A_1,A_2,\cdots,A_N of length 3 each, using each of the integers 1 through 3N exactly once.
* Let P be an empty sequence, and do the following operation 3N times.
* Among the elements that are at the beginning of one of the sequences A_i that is non-empty, let the smallest be x.
* Remove x from the sequence, and add x at the end of P.

Constraints

* 1 \leq N \leq 2000
* 10^8 \leq M \leq 10^9+7
* M is a prime number.
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N M


Output

Print the number of permutations modulo M.

Examples

Input

1 998244353


Output

6


Input

2 998244353


Output

261


Input

314 1000000007


Output

182908545
// No.213741032120
#include <bits/stdc++.h>
#define N 2020
using namespace std;

int n, eps, Mod, f[N * 3][N * 4];

inline void inc(int &x, const long long &y) {
    return x = (x + y) % Mod, void();
}

int main() {
#ifdef isLOCAL
    freopen(".in", "r", stdin);
    freopen(".out", "w", stdout);
#endif

    cin >> n >> Mod;
    eps = n + 1;
    f[0][eps] = 1;

    for (int i = 0; i < n * 3; ++i)
        for (int j = -n, k; j <= 3 * n; ++j) if ((k = f[i][j + eps])) {
            inc(f[i + 1][j + 1 + eps], k);
            inc(f[i + 2][j - 1 + eps], 1ll * k * (i + 1));
            inc(f[i + 3][j + eps], 1ll * k * (i + 1) * (i + 2));
        }

    int ans = 0;
    for (int i = 0; i <= 3 * n; i += 3)
        ans = (ans + f[n * 3][i + eps]) % Mod;
    cout << ans << endl;

    return 0;
}
