Farmer John has come up with a new morning exercise routine for the cows (again)!
As before, Farmer John's N cows (1≤N≤7500) are standing in a line. The i-th cow from the left has label i for each 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 Ai-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 (108≤M≤109+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
}
INPUT FORMAT (file exercise.in):
The first line contains N and M.
OUTPUT FORMAT (file exercise.out):
A single integer.
SAMPLE INPUT:
5 1000000007
SAMPLE OUTPUT:
369329541
For each 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\cdot 2^{25}\cdot 3^{20}\cdot 4^{30}\cdot 5^{24}\cdot 6^{20}\equiv 369329541\pmod{10^9+7}.
Note: This problem has an expanded memory limit of 512 MB.
SCORING:
- Test case 2 satisfies N=8.
- Test cases 3-5 satisfy N\le 50.
- Test cases 6-8 satisfy N\le 500.
- Test cases 9-12 satisfy N\le 3000.
- Test cases 13-16 satisfy no additional constraints.
Problem credits: Benjamin Qi