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 int LL;

#define st first
#define nd second
#define PLL pair <LL, LL>
#define PII pair <int, int>

const int N = 2007;

int n, mod;
int newt[3 * N][3 * N];
int dp1[3 * N][2 * N];
int dp2[3 * N][2 * N];

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

int main(){
	scanf("%d %d", &n, &mod);
	for(int i = 0; i <= 3 * n; ++i){
		newt[i][0] = newt[i][i] = 1;
		for(int j = 1; j < i; ++j)
			newt[i][j] = (newt[i - 1][j] + newt[i - 1][j - 1]) % mod;
	}

	dp1[0][0] = 1;
	for(int i = 1; i <= 3 * n; ++i)
		for(int j = 0; j <= 2 * n; ++j){
			add(dp1[i][j], 1LL * dp1[i - 1][j + 1] * (j + 1) % mod);
			if(j > 1)	add(dp1[i][j], dp1[i - 1][j - 2]);
		}
	
	dp2[0][0] = 1;
	for(int i = 1; i <= 3 * n; ++i)
		for(int j = 0; j <= 2 * n; ++j){
			add(dp2[i][j], 1LL * dp2[i - 1][j + 1] * (j + 1) % mod);
			if(j > 0)	add(dp2[i][j], dp2[i - 1][j - 1]);
		}
	
	int ans = 0;
	for(int a = 0; a <= n; ++a)
		for(int b = 0; b <= n; ++b)
			if(a + b <= n){
				LL tmp = 1LL * newt[3 * n][3 * a] * dp1[3 * a][0] % mod;
				tmp = tmp * newt[3 * n - 3 * a][2 * b] % mod * dp2[2 * b][0] % mod;
//				printf("%d %d -> %d\n", a, b, tmp);
				add(ans, tmp);
			}
	
	printf("%d\n", ans);
	return 0;
}