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;
int mod,n;
const int N = 1e4+5;
int fac[N], ifac[N], h[N], g[N];
inline int qpow(int a,int b){int ret=1;for(;b;b>>=1,a=1ll*a*a%mod)if(b&1)ret=1ll*ret*a%mod;return ret;}

inline void init(int n){
	fac[0]=ifac[0]=1;for(int i=1;i<=n;i++)fac[i]=1ll*fac[i-1]*i%mod;
	ifac[n]=qpow(fac[n],mod-2);for(int i=n-1;i;i--)ifac[i]=1ll*ifac[i+1]*(i+1)%mod;
	g[0]=1;for(int i=1;i<=n;i++)g[i]=1ll*g[i-1]*ifac[3]%mod*2%mod;
	h[0]=1;for(int i=1;i<=n;i++)h[i]=1ll*h[i-1]*ifac[2]%mod;
}
int ans=0;

int main(){
	cin >> n >> mod;
	init(n*3);
	int ans=0;
	for(int i=0;i<=n*3;i++){
		for(int j=0;j*2+i<=n*3&&j<=i;j++){
			if((i+2*j)%3==0){
				int k=(3*n-i-2*j)/3;
				ans = (1ll*ans + 1ll*h[j]*g[k]%mod*ifac[i]%mod*ifac[j]%mod*ifac[k]%mod*fac[n*3]%mod)%mod;
			}
		}
	}
	cout << ans << endl;
}
