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;

int n, m;
int mem[4010][6010][3];

void add(int &x, int v)
{
    x = (x + v) % m;
}

int main()
{
    scanf("%d%d", &n, &m);

    mem[1][1][0] = 1;
    for(int i = 1; i <= 2 * n; i++)
    {
        for(int j = 1; j <= 3 * n - 1; j++)
        {
            for(int k = 0; k < 3; k++)
            {
                if(k == 0)
                    add(mem[i][j + 1][0], mem[i][j][k]);
                else
                    add(mem[i + 1][j + 1][0], mem[i][j][k]);

                if(k < 2)
                    add(mem[i][j + 1][k + 1], 1LL * mem[i][j][k] * j % m);
            }
        }
    }
    
    int r = 0;
    for(int i = 1; i <= n; i++)
        for(int k = 0; k < 3; k++)
            add(r, mem[i][3 * n][k]);
    add(r, mem[n + 1][3 * n][0]);

    printf("%d\n", r);
    return 0;
}
