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;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)

int main(){
  int N, MOD; cin >> N >> MOD;
  vector<vector<long long>> dp(3 * N + 1, vector<long long>(6 * N + 3, 0));
  dp[0][N * 3 + 1] = 1;
  rep(i, N * 3 + 1) rep(k, 6 * N) {
    if (i == 0 || k == 0) continue;
    long long num = 0;
    if (i >= 3) { num += dp[i - 3][k] * (3 * N - i + 2) * (3 * N - i + 1); num %= MOD; }
    if (i >= 2) { num += dp[i - 2][k - 1] * (3 * N - i + 1); num %= MOD; }
    if (i >= 1) { num += dp[i - 1][k + 1]; num %= MOD; }
    dp[i][k] = num;
  }
  long long ans = 0;
  rep(i, N * 3 + 2) {
    ans += dp[3 * N][i];
    ans %= MOD;
  }
  cout << ans << endl;
}
