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: Time Limit Exceeded)

pub fn main() {
    use std::collections::HashMap;
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    let n: usize = s.trim().parse().unwrap();
    s.clear();
    std::io::stdin().read_line(&mut s).unwrap();
    let a: Vec<u64> = s.split_whitespace().map(|w| w.parse().unwrap()).collect();
    let modular: u64 = 1_000_000_007;
    let mut factors: Vec<HashMap<u64, u32>> = vec![HashMap::new(); n];
    for i in 0..n {
        let mut x = a[i];
        let mut d = 2;
        while x > 1 && d * d <= x {
            if x % d == 0 {
                let mut e = 0;
                while {
                    x /= d;
                    e += 1;
                    x % d == 0
                } {}
                factors[i].insert(d, e);
            }
            d += 1;
        }
        if x > 1 {
            factors[i].insert(x, 1);
        }
    }
    let mut lcm_factors: HashMap<u64, u32> = HashMap::new();
    for i in 0..n {
        for (&p, &e) in &factors[i] {
            let k = lcm_factors.entry(p).or_insert(0);
            *k = std::cmp::max(*k, e);
        }
    }
    let mut ans = 0;
    for i in 0..n {
        let mut b = 1;
        for (&p, &k) in &lcm_factors {
            let e = *factors[i].get(&p).unwrap_or(&0);
            b = b * p.pow(k - e) % modular;
        }
        ans = (ans + b) % modular;
    }
    println!("{}", ans);
}