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 (1.42.0) (judge verdict: Wrong Answer)

use std::io;
use std::str::FromStr;

static MOD: u128 = 1_000_000_007;

fn input<T:FromStr>() -> T {
    let mut line: String = String::new();
    io::stdin().read_line(&mut line);
    line.trim().parse().ok().unwrap()
}

fn input_vec() -> Vec<u128> {
    let mut line: String = String::new();
    io::stdin().read_line(&mut line);
    line.split_whitespace().map(|s| s.parse().unwrap()).collect()
}

fn gcd(a: u128, b: u128) -> u128 {
    if b == 0 {
      a
    } else {
      gcd(b, a % b)
    }
}

fn binary_power(x: u128, n: u128) -> u128 {
    if n == 0 {
      return 1;
    }
    if n % 2 == 0 {
      return (binary_power(x, n / 2) % MOD).pow(2) % MOD;
    } else {
      return binary_power(x, n - 1) % MOD * x % MOD;
    }
}
  
fn inverse_mod(n: u128) -> u128 {
    binary_power(n, MOD-2)
} 

fn main() {
    let n: usize = input();
    let a_vec = input_vec();
    let mut lcm = 1;
    for a in &a_vec {
        // lcm = *a * lcm % MOD * inverse_mod(gcd(lcm, *a)) % MOD;
        lcm = *a * lcm * inverse_mod(gcd(lcm, *a)) % MOD;
        // println!("{}", lcm);
    }
    // println!("{}", lcm % MOD);
    let mut sum = 0;
    for a in a_vec {
        sum += lcm * inverse_mod(a) % MOD;
        sum %= MOD;
    }
    println!("{}", sum % MOD);
}