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: Wrong Answer)

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();

    Map<Integer, Integer> lcm = f(A[0]);
    for (int i = 1; i < N; i++)
      lcm = lcm(lcm, f(A[i]));
    System.out.println(lcm);

    long mod = 1000000007;
    long ans = 0;
    for (int i = 0; i < N; i++) {
      Map<Integer, Integer> Bi = divide(lcm, f(A[i]));
    System.out.println(f(A[i]));
    System.out.println(Bi);
      long b = 1;
      for (Integer p : Bi.keySet()) {
        for (int j = 0; j < Bi.get(p); j++) {
          b *= p;
          b %= mod;
        }
      }
      ans += b;
      ans %= mod;
    }
    System.out.println(ans);
  }

  private static Map<Integer, Integer> divide(Map<Integer, Integer> a, Map<Integer, Integer> b) {
    Map<Integer, Integer> r = new HashMap<>();
    Set<Integer> ps = new HashSet<>();
    ps.addAll(a.keySet());
    ps.addAll(b.keySet());
    for (Integer p : ps) {
      int av = a.getOrDefault(p, 0);
      int bv = b.getOrDefault(p, 0);
      r.put(p, av - bv);
    }
    return r;
  }

  private static Map<Integer, Integer> lcm(Map<Integer, Integer> a, Map<Integer, Integer> b) {
    Map<Integer, Integer> r = new HashMap<>();
    Set<Integer> ps = new HashSet<>();
    ps.addAll(a.keySet());
    ps.addAll(b.keySet());
    for (Integer p : ps) {
      int av = a.getOrDefault(p, 0);
      int bv = b.getOrDefault(p, 0);
      r.put(p, Math.max(av, bv));
    }
    return r;
  }

  private static Map<Integer, Integer> f(int n) {
    Map<Integer, Integer> f = new HashMap<>();
    int m = n;
    int s = SIEVE[m - 1];
    while (s != 0) {
      f.compute(s, (k, v) -> v == null ? 1 : v + 1);
      m /= s;
      s = SIEVE[m - 1];
    }
    return f;
  }

  private static int[] SIEVE = sieve(1000000);

  private static int[] sieve(int n) {
    int [] sieve = new int[n];
    for (int i = 2; i <= n; i += 2) sieve[i - 1] = 2;
//    int sqrt_n = (int) Math.sqrt(n);
    for (int i = 3; i <= n; i += 2) {
      if(sieve[i - 1] == 0) sieve[i - 1] = i;
      for (int j = i + i; j <= n; j += i)
        if(sieve[j - 1] == 0) sieve[j - 1] = i;
    }
    return sieve;
  }
}