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)

use std::io::{self, BufRead};
use std::collections::HashMap;

const M: i64 = 1000000007;

pub fn find_prime_factors(n: i64) -> HashMap<i64, i64> {
  let mut res = HashMap::new();
  let mut i = 2;
  let mut m = n;
  while i * i <= m {
    while m % i == 0 {
      let val = 1 + res.get(&i).unwrap_or(&0);
      res.insert(i, val);
      m /= i;
    }
    i += 1
  }
  if m != 1 {
    let val = 1 + res.get(&m).unwrap_or(&0);
    res.insert(m, val);
  }
  res
}

fn main() {
  let stdin = io::stdin();
  let mut lines = stdin.lock().lines();
  let _ = lines.next();
  let line = lines.next().unwrap().unwrap();
  let mut max_x: HashMap<i64, i64> = HashMap::new();
  let xs: Vec<_> = line.split(' ').map(|x| x.parse().unwrap()).map(|x| {
    let x = find_prime_factors(x);
    for (p, n) in x.iter() {
      if max_x.contains_key(p) {
        let m = max_x.get(p).map(|m| *m).unwrap();
        if m < *n {
          max_x.insert(*p, *n);
        }
      } else {
        max_x.insert(*p, *n);
      }
    }
    x
  }).collect();
  let ans = xs.iter().map(|x| {
    let mut z = 1;
    for (p, n) in max_x.iter() {
      let n0 = x.get(p).map(|p| *p).unwrap_or(0);
      let l = *n - n0;
      for _ in 0..l {
        z = (z * p) % M;
      }
    }
    z
  }).fold(0, |acc, z| (acc + z) % M);
  println!("{}", ans);
}