Exercise (Platinum)
Farmer John has come up with a new morning exercise routine for the cows (again)!
As before, Farmer John's n cows are standing in a line. The i-th cow from the left has label i for each i (1 ≤ i ≤ n). He tells them to repeat the following step until the cows are in the same order as when they started.
Given a permutation A of length n, the cows change their order such that the i-th cow from the left before the change is A[i]-th from the left after the change.
For example, if A = (1, 2, 3, 4, 5) then the cows perform one step and immediately return to the same order. If A = (2, 3, 1, 5, 4), then the cows perform six steps before returning to the original order. The order of the cows from left to right after each step is as follows:
0 steps: (1, 2, 3, 4, 5)
1 step: (3, 1, 2, 5, 4)
2 steps: (2, 3, 1, 4, 5)
3 steps: (1, 2, 3, 5, 4)
4 steps: (3, 1, 2, 4, 5)
5 steps: (2, 3, 1, 5, 4)
6 steps: (1, 2, 3, 4, 5)
Compute the product of the numbers of steps needed over all n! possible permutations A of length n.
As this number may be very large, output the answer modulo m.
Contestants using C++ may find the following code from KACTL helpful. Known as the Barrett reduction, it allows you to compute a % b several times faster than usual, where b > 1 is constant but not known at compile time. (we are not aware of such an optimization for Java, unfortunately).
#include <bits/stdc++.h>
using namespace std;
typedef unsigned long long ull;
typedef __uint128_t L;
struct FastMod {
ull b, m;
FastMod(ull b) : b(b), m(ull((L(1) << 64) / b)) {}
ull reduce(ull a) {
ull q = (ull)((L(m) * a) >> 64);
ull r = a - q * b; // can be proven that 0 <= r < 2*b
return r >= b ? r - b : r;
}
};
FastMod F(2);
int main() {
int M = 1000000007; F = FastMod(M);
ull x = 10ULL*M+3;
cout << x << " " << F.reduce(x) << "\n"; // 10000000073 3
}Input
The first line contains n (1 ≤ n ≤ 7500) and m (10^8 ≤ m ≤ 10^9 + 7, m is prime).
Output
Print a single integer.
Example
For each i (1 ≤ i ≤ n), the i-th element of the following array is the number of permutations that cause the cows to take i steps: [1, 25, 20, 30, 24, 20]. The answer is 1^1 * 2^25 * 3^20 * 4^30 * 5^24 * 6^20 = 369329541 (mod 10^9 + 7).