Problem:
Score : 500 points

Problem Statement

Given are N positive integers A_1,...,A_N.

Consider positive integers B_1, ..., B_N that satisfy the following condition.

Condition: For any i, j such that 1 \leq i < j \leq N, A_i B_i = A_j B_j holds.

Find the minimum possible value of B_1 + ... + B_N for such B_1,...,B_N.

Since the answer can be enormous, print the sum modulo (10^9 +7).

Constraints

1 \leq N \leq 10^4

1 \leq A_i \leq 10^6

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 ... A_N

Output

Print the minimum possible value of B_1 + ... + B_N for B_1,...,B_N that satisfy the condition, modulo (10^9 +7).

Sample Input 1

3
2 3 4

Sample Output 1

13

Let B_1=6, B_2=4, and B_3=3, and the condition will be satisfied.

Sample Input 2

5
12 12 12 12 12

Sample Output 2

5

We can let all B_i be 1.

Sample Input 3

3
1000000 999999 999998

Sample Output 3

996989508

Print the sum modulo (10^9+7).

Solution in Rust (1.15.1) (judge verdict: Wrong Answer)

fn read<T: std::str::FromStr>(
    stdin_lock: &mut std::io::StdinLock,
    buf: &mut Vec<u8>,
    delimiter: u8,
) -> T {
    buf.clear();
    let l = std::io::BufRead::read_until(stdin_lock, delimiter, buf).unwrap();
    buf.truncate(l - 1); // remove delimiter
    let s = unsafe { std::str::from_utf8_unchecked(&buf) };
    s.parse().unwrap_or_else(|_| panic!("read"))
}

fn gcd(a: u64, b: u64) -> u64 {
    if b == 0 {
        a
    } else {
        gcd(b, a % b)
    }
}

fn lcm(a: u64, b: u64) -> u64 {
    a * b / gcd(a, b)
}

fn main() {
    let stdin = std::io::stdin();
    let mut stdin_lock = stdin.lock();
    let mut buf: Vec<u8> = Vec::new();
    let n: usize = read(&mut stdin_lock, &mut buf, b'\n');
    let mut av = vec![0u64; n];
    for i in 0..n - 1 {
        av[i] = read(&mut stdin_lock, &mut buf, b' ');
    }
    av[n - 1] = read(&mut stdin_lock, &mut buf, b'\n');

    let mut l = 1;
    for i in 0..av.len() {
        l = lcm(av[i], l);
    }

    let mut bv = vec![0u64; n];
    for i in 0..av.len() {
        bv[i] = l / av[i];
    }

    let mut ans = 0;
    let m = 1_000_000_007;
    for i in 0..bv.len() {
        ans = ans + bv[i];
    }
    println!("{}", ans % m);
}