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
// -----------------------------------
// Author     : MatsuTaku
// Affiliation: Tokushima University
// Country    : Japan
// Date       : 03/21/2020
// -----------------------------------

#include <bits/stdc++.h>

using namespace std;
using ll = long long;

int main() {
  cin.tie(nullptr); ios::sync_with_stdio(false);

  ll N,M; cin>>N>>M;

  vector<vector<ll>> dp(3*N+3, vector<ll>(N+2, 0));
  dp[0][0] = 1;
  for (int i = 0; i < 3*N; i++) {
    for (int j = 0; j <= N; j++) {
      dp[i+1][j] += dp[i][j];
      dp[i+1][j] %= M;
      dp[i+2][j+1] += dp[i][j]*(i+1)%M;
      dp[i+2][j+1] %= M;
      dp[i+3][j+1] += dp[i][j]*(i+2)*(i+1)%M;
      dp[i+3][j+1] %= M;
    }
  }

  ll ans = 0;
  for (int i = 0; i <= N; i++) {
    ans += dp[3*N][i];
    ans %= M;
  }
  cout << ans << endl;

  return 0;
}
