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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.math.BigInteger;
import java.util.*;

public class Main{
  public static long inved(long a, long modulo){
    long x = 1, y = 0, u = 0, v = 1, k = a, l = modulo;
    long tmp;
    while (l != 0){
      tmp = x;
      x = u;
      u = tmp - u * (k / l);
      tmp = y;
      y = v;
      v = tmp - v * (k / l);
      tmp = k;
      k = l;
      l = tmp % l;
    }
    return (x + modulo) % modulo;
  }
  public static BigInteger gcd(BigInteger a, BigInteger b){
    BigInteger x = a, y = b;
    BigInteger tmp;
    while (y != BigInteger.ZERO){
      tmp = x;
      x = y;
      y = tmp.mod(y);
    }
    return x;
  }
  public static void main(String[] args){
    Scanner sc = new Scanner(System.in);
    long modulo = (long)1e9 + 7;
    int n = sc.nextInt();
    long a[] = new long[n];
    BigInteger L = BigInteger.ONE;
    BigInteger t;
    for (int i=0;i<n;i++){
      a[i] = sc.nextLong();
      t = gcd(L, BigInteger.valueOf(a[i]));
      t = BigInteger.valueOf(a[i]).divide(t);
      L = L.multiply(t);
    }
    long S = 0l;
    for (int i=0;i<n;i++){
      S += inved(a[i], modulo);
      S %= modulo;
    }
    L = L.multiply(BigInteger.valueOf(S));
    System.out.println(L.mod(BigInteger.valueOf(modulo)));
  }
}