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 <algorithm>
#include <iomanip>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;
typedef long long ll;
ll N, M;
ll dp[6005][2005];

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout << setprecision(10) << fixed;
    cin >> N >> M;
    dp[3*N+1][0] = 1;
    for(int i = 3*N+1; i >= 1; i--){
        for(int j = 0; j <= N; j++){
            dp[i-1][j] += dp[i][j];
            dp[i-1][j] %= M;
            if(j < N){
                if(i-2 >= 0){
                    ll c = i-2;
                    dp[i-2][j+1] += c*dp[i][j];
                    dp[i-2][j+1] %= M;
                }
                if(i-3 >= 1){
                    ll c = ((i-2)*(i-3))%M;
                    dp[i-3][j+1] += c*dp[i][j];
                    dp[i-3][j+1] %= M;
                }
            }
        }
    }
    ll ans = 0;
    for(int i = 0; i <= N; i++){
        ans += dp[1][i];
        ans %= M;
    }
    cout << ans << endl;
}