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;

const int Maxn = 6005;
const int Maxm = 2005;

int n, mod;
int dp[Maxn][Maxm];

int main()
{
    scanf("%d %d", &n, &mod);
    dp[0][0] = 1;
    for (int i = 1; i <= 3 * n; i++)
        for (int j = 0; j <= n; j++) {
            dp[i][j] = (dp[i][j] + dp[i - 1][j]) % mod;
            if (i >= 2 && j >= 1)
                dp[i][j] = (dp[i][j] + ll(dp[i - 2][j - 1]) * (i - 1)) % mod;
            if (i >= 3 && j >= 1)
                dp[i][j] = (dp[i][j] + ll(dp[i - 3][j - 1]) * (i - 1) % mod * (i - 2)) % mod;
        }
    int res = 0;
    for (int j = 0; j <= n; j++)
        res = (res + dp[3 * n][j]) % mod;
    printf("%d\n", res);
    return 0;
}
