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;

#define Fi first
#define Se second
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
#define all(x) (x).begin(), (x).end()
#define pb push_back
#define szz(x) (int)(x).size()
#define rep(i, n) for(int i=0;i<(n);i++)
typedef tuple<int, int, int> t3;

ll mod;
int N;
int dp[6060][10010];
int base = 3030;

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