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 <cstdio>

const int MAXN = 6000;

int f[4][2*MAXN + 5], N, M;

int main() {
	scanf("%d%d", &N, &M);
	
	f[0][MAXN] = 1;
	for(int i=1;i<=3*N;i++) {
		int lb = MAXN - (i - 1), ub = MAXN + (i - 1);
		for(int j=lb;j<=ub;j++) {
			for(int o=3;o>=1;o--)
				f[o][j] = f[o-1][j];
			f[0][j] = 0;
		}
		for(int j=lb;j<=ub;j++) {
			f[0][j+1] = (f[0][j+1] + f[1][j]) % M;
			int del = (i - 1);
			f[0][j-1] = (f[0][j-1] + 1LL*f[2][j]*del%M) % M;
			del = 1LL*(i - 1)*(i - 2)%M;
			f[0][j] = (f[0][j] + 1LL*f[3][j]*del%M) % M;
		}
	}
	int ans = 0;
	for(int i=MAXN;i<=MAXN+3*N;i++)
		ans = (ans + f[0][i]) % M;
	printf("%d\n", ans);
}