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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.math.*;
import java.util.*;
import java.util.function.*;
 
public class Main {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int N = sc.nextInt();
    int[] A = new int[N];
    for (int i = 0; i < N; i++) {
      A[i] = sc.nextInt();
    }
    sc.close();

//    BigInteger lcm = Arrays.stream(A).parallel()
//        .mapToObj(BigInteger::valueOf)
//        .reduce(BigInteger.ONE, (a, b) -> a.divide(a.gcd(b)).multiply(b));

    BigInteger lcm = BigInteger.ONE;
    for (int ai : A) {
      BigInteger a = BigInteger.valueOf(ai);
      lcm = lcm.divide(lcm.gcd(a)).multiply(a);
    }

    long mod = 1000000007;
    long modLcm = lcm.remainder(BigInteger.valueOf(mod)).longValue();

    long ans = 0;
    for (int ai : A)
      ans = (ans + modDiv(modLcm, ai, mod)) % mod;
    System.out.println(ans);
  }

  private static long modDiv(long a, long b, long mod) {
    return (a % mod) * modInv(b, mod) % mod;
  }

  private static long modInv(long x, long mod) {
    return modPow(x, mod - 2, mod);
  }

  private static long modPow(long b, long e, long mod) {
    long ans = 1;
    long x = b % mod;
    for (long y = e; y != 0; y >>= 1) {
      if ((y & 1) == 1) ans = (ans * x) % mod;
      x = (x * x) % mod;
    }
    return ans;
  }
}