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 <iostream>
#include <vector>

using namespace std;

int main(){
    int N, M; cin >> N >> M;
    vector<vector<long long>> dp(3*N+1, vector<long long>(N+1, 0));
    dp[3*N][0] = 1;
    for(int i=3*N;i>0;i--){
        for(int j=0;j<=N;j++){
            long long m = 1;
            for(int k=1;k<=3;k++){
                if(i-k < 0) break;
                if(j+k/2 > N) break;
                dp[i-k][j+k/2] += dp[i][j] * m;
                dp[i-k][j+k/2] %= M;
                m = m * (i-k) % M;
            }
        }
    }
    long long res = 0;
    for(int i=0;i<=N;i++) res = (res + dp[0][i]) % M;
    cout << res << endl;
}