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;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return true; } return false; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return true; } return false; }
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define si(x) int(x.size())
const int mod=1000000007,MAX=6005,INF=1<<29;
ll dp[MAX][2*MAX];

int main(){
    
    std::ifstream in("text.txt");
    std::cin.rdbuf(in.rdbuf());
    cin.tie(0);
    ios::sync_with_stdio(false);
    
    int N,M;cin>>N>>M;
    N*=3;
    dp[0][6000]=1;
    
    for(int i=0;i<N;i++){
        for(int j=1;j<=12000;j++){
            dp[i+3][j]+=dp[i][j]*(N-i-1)%M*(N-i-2)%M;
            dp[i+3][j]%=M;
            
            dp[i+2][j-1]+=dp[i][j]*(N-1-i)%M;
            dp[i+2][j-1]%=M;
            
            dp[i+1][j+1]+=dp[i][j];
            dp[i+1][j+1]%=M;
        }
    }
    
    ll ans=0;
    
    for(int j=6000;j<=12000;j++){
        ans+=dp[N][j];
        ans%=M;
    }
    
    cout<<ans<<endl;
}
