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>
typedef long long int ll;
typedef long double ld;
#define pb push_back
#define pii pair < int, int >
#define F first
#define S second
#define int long long int
#define sync ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#pragma GCC optimize ("Ofast")
#pragma GCC optimize ("unroll-loops")
#define endl '\n'
#define kill(x) return cout<<x<<'\n', 0;
using namespace std;
const int N=2e3+10;
ll mod;
ll dp[N*3][N];
ll ent(ll k){
    return k*(k-1)/2%mod;
}
int32_t main(){
    sync;
    ll n;
    cin >> n;
    cin >> mod;
    ll a=3*n;
    dp[0][0]=1;
    dp[1][0]=1;
    dp[2][0]=1;
    dp[2][1]=1;
    for (int i=3;i<=a;i++){
        for (int j=0;j<N;j++){
            dp[i][j]+=dp[i-1][j];
            if (j>0){
                dp[i][j]+=dp[i-2][j-1]*(i-1)%mod;
                dp[i][j]%=mod;
                dp[i][j]+=2*dp[i-3][j-1]*ent(i-1)%mod;
                dp[i][j]%=mod;
            }
        }
    }
    ll ans=0;
    for (int i=0;i<=n;i++){
        ans+=dp[a][i];
    }
    cout << ans%mod;
}
