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>
#define ll long long
#define N 6005
using namespace std;

int n;
ll P, f[N][N*2];

void add(ll &a, ll b){
    a += b;
    if(a >= P) a -= P;
}

int main()
{
    cin >> n >> P;
    f[0][N] = 1;
    n *= 3;
    for(int i = 0; i <= n; i++){
        for(int j = -i; j <= i; j++){
            // 考虑从后往前逐位确定,除了每段的开头,位置x的方案为x
            // 但每段开头必须是前缀最大值,所以只有一种方案
            add(f[i+1][j+N+1], f[i][j+N]);
            add(f[i+2][j+N-1], f[i][j+N]*(i+1)%P);
            add(f[i+3][j+N], f[i][j+N]*(i+1)%P*(i+2)%P);
        }
    }
    ll ans = 0;
    for(int i = 0; i <= n; i++){
        add(ans, f[n][i+N]);
    }
    cout << ans << endl;

    return 0;
}