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
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> l_l;
typedef pair<int, int> i_i;
template<class T>
inline bool chmax(T &a, T b) {
    if(a < b) {
        a = b;
        return true;
    }
    return false;
}

template<class T>
inline bool chmin(T &a, T b) {
    if(a > b) {
        a = b;
        return true;
    }
    return false;
}

ll mod = 1000000007;
ll N;
ll Prev[6005], Next[6005];
ll ans;

int main() {
    cin >> N >> mod;
    for(ll i = 0; i <= 3 * N; i++) {
        Prev[i] = 1;
    }
    ans = 1;
    for(int timer = 0; timer < N; timer++) {
        for(ll rest = 0; rest <= 3 * N; rest++) Next[rest] = 0;
        for(ll rest = 0; rest <= 3 * N; rest++) {
            if(rest >= 2) {
                Next[rest-2] += Prev[rest] * (rest - 1);
                Next[rest-2] %= mod;
            }
            if(rest >= 3) {
                Next[rest-3] += Prev[rest] * (rest - 1) * (rest - 2);
                Next[rest-3] %= mod;
            }
        }
        for(ll rest = 3 * N - 1; rest >= 0; rest--) {
            Next[rest] += Next[rest+1];
            Next[rest] %= mod;
        }
        ans += Next[0];
        ans %= mod;
        swap(Prev, Next);
    }
    cout << ans << endl;
    return 0;
}