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;
int n, m, ans;
const int M = 6001;
int dp[6005][12005];
int inc(int x, int y){
    return x + y >= m ? x + y - m : x + y;
}
int main(){
    cin >> n >> m; n = n * 3;
    dp[0][M] = 1;
    for(int i = 0; i <= n; i++)
        for(int j = -i; j <= i; j++){
            dp[i + 1][j + 1 + M] = inc(dp[i + 1][j + M + 1], dp[i][j + M]);
            dp[i + 2][j - 1 + M] = inc(dp[i + 2][j + M - 1], 1ll * dp[i][j + M] * (i + 1) % m);
            dp[i + 3][j + M] = inc(dp[i + 3][j + M], 1ll * dp[i][j + M] * (i + 1) % m * (i + 2) % m); 
        } 
   //cout << dp[3][0 + M] << endl;
   for(int i = 0; i <= n ; i++){
        ans = inc(ans, dp[n][i + M]);
    }
    cout << ans << endl;
}