Exercise

Farmer John has come up with a new morning exercise routine for the cows (again)!

As before, Farmer John's $N$ cows ($1\le N\le 7500$) are standing in a line. The $i$-th cow from the left has label $i$ for each $1\le i\le N$. 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 $M$ ($10^8\le M\le 10^9+7$, $M$ is prime).

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
}

For each $1\le i\le N$, 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 $1^1\cdot 2^{25}\cdot 3^{20}\cdot 4^{30}\cdot 5^{24}\cdot 6^{20}\equiv 369329541(\mathrm{mod}10^9+7)$.

Note: This problem has an expanded memory limit of 512 MB.

Problem credits: Benjamin Qi

Source: USA Computing Olympiad.

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