Return the number of permutations of 1 to n so that prime numbers are at prime indices (1-indexed.)
(Recall that an integer is prime if and only if it is greater than 1, and cannot be written as a product of two positive integers both smaller than it.)
Since the answer may be large, return the answer modulo 10^9 + 7.
Example 1:
Input: n = 5 Output: 12 Explanation: For example [1,2,5,4,3] is a valid permutation, but [5,2,3,4,1] is not because the prime number 5 is at index 1.
Example 2:
Input: n = 100 Output: 682289015
Constraints:
1 <= n <= 100program main
    implicit none
    integer :: i, j, k, l, m, n, p, q, r, s, t, u, v, w, x, y, z
    integer :: count, modulus
    logical :: is_prime(100)
    ! Modulus for the result
    modulus = 1000000007
    ! Initialize the array of primes
    do i = 2, 100
        is_prime(i) = .true.
    end do
    ! Mark non-primes
    do i = 2, 100
        if (is_prime(i)) then
            j = i * i
            do while (j <= 100)
                is_prime(j) = .false.
                j = j + i
            end do
        end if
    end do
    ! Count the number of permutations
    count = 0
    do i = 1, 100
        if (is_prime(i)) then
            count = count + 1
        end if
    end do
    ! Output the result modulo the modulus
    print '(I0)', count
end program main25
```
program prime_permutation
    implicit none
    integer, parameter :: modulus = 1000000007
    integer :: n, i, j, count
    logical, allocatable :: is_prime(:)
    ! read input
    read (*,*) n
    ! allocate array for prime numbers
    allocate (is_prime(n))
    ! initialize array with true
    is_prime = .true.
    ! mark non-prime numbers as false
    do i = 2, n
        if (is_prime(i)) then
            do j = i * i, n, i
                is_prime(j) = .false.
            end do
        end if
    end do
    ! count prime permutations
    count = 0
    do i = 1, n
        if (is_prime(i)) then
            count = count + 1
        end if
    end do
    ! output result modulo 10^9 + 7
    write (*,*) count
    ! deallocate array
    deallocate (is_prime)
end program prime_permutation
```
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          def numPrimeArrangements(n: int) -> int:
    def is_prime(num):
        if num < 2:
            return False
        for i in range(2, int(num ** 0.5) + 1):
            if num % i == 0:
                return False
        return True
    prime_count = sum(1 for i in range(2, n + 1) if is_prime(i))
    composite_count = n - prime_count
    MOD = 10**9 + 7
    res = 1
    for i in range(1, prime_count + 1):
        res = res * i % MOD
    for i in range(1, composite_count + 1):
        res = res * i % MOD
    return res
The algorithm calculates the total number of primes and composites up to the given integer n. Then, it multiplies the factorials of the two counts (mod 10^9 + 7), eventually returning the result as the number of permutations of 1 to n where prime numbers are at the prime indices.
#include <vector>
using namespace std;
const int MOD = 1e9 + 7;
int numPrimeArrangements(int n) {
    vector<bool> is_prime(n + 1, true);
    is_prime[0] = false;
    is_prime[1] = false;
    for (int i = 2; i * i <= n; i++) {
        if (is_prime[i]) {
            for (int j = i * i; j <= n; j += i) {
                is_prime[j] = false;
            }
        }
    }
    int prime_count = 0;
    for (int i = 2; i <= n; i++) {
        if (is_prime[i]) prime_count++;
    }
    int composite_count = n - prime_count;
    long long res = 1;
    for (int i = 1; i <= prime_count; i++) {
        res = res * i % MOD;
    }
    for (int i = 1; i <= composite_count; i++) {
        res = res * i % MOD;
    }
    return res;
}
The algorithm calculates the total number of primes and composites up to the given integer n. Then, it multiplies the factorials of the two counts (mod 10^9 + 7), eventually returning the result as the number of permutations of 1 to n where prime numbers are at the prime indices.