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)

macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

macro_rules! stdin {
    () => {{
        use std::io::Read;
        let mut s = String::new();
        std::io::stdin().read_to_string(&mut s).unwrap();
        s
    }};
}

macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

// Test Code
macro_rules! test {
    ($($input:expr => $output:expr),* $(,)*) => {
        #[test]
        fn solve_test() {
            $(
                assert_eq!(solve($input), $output);
            )*
        }
    };
}

test! {
}

fn main() {
    println!("{}", solve(&stdin!()));
}

fn solve(src: &str) -> String {
    input!{
        source = src,
        n: usize,
        an: [u64; n],
    }

    let g = mul_lcd(&an);

    let bn = an.iter().map(|a| g / a );

    let mut r : u64 = 0;
    for b in bn {
        r += b;
    }
    return (r % 1000000007).to_string();
}

fn gcd(a: u64, b: u64) -> u64 {
    let mut x = a;
    let mut y = b;
    while y != 0 {
        let t = y;
        y = x % y;
        x = t;
    }
    x
}

fn mul_lcd(ns: &Vec<u64>) -> u64 {
    let mut lcd = ns[0];
    for n in ns.iter() {
        let g = gcd(*n, lcd);
        lcd = lcd * n / g;
    }
    return lcd;
}