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: Runtime Error)

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.List;
import java.util.Map.Entry;
import java.util.PriorityQueue;
import java.util.Scanner;

class Main {

    static final int MOD = 1000000007;
    static final int INF = 1 << 30;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        long ans = 0;
        int N = sc.nextInt();
        int[] a = new int[N];
        int max = 0;
        for (int i = 0; i < N; i++) {
            a[i] = sc.nextInt();
            max = Math.max(max, a[i]);
        }
        int[] lcm = new int[max + 1];
        List<Integer> prime = primeList(max);
        for (int i = 0; i < N; i++) {
            int n = a[i];
            for (int p : prime) {
                int cnt = 0;
                while (n % p == 0) {
                    cnt++;
                    n /= p;
                }
                lcm[p] = Math.max(lcm[p], cnt);
                if (p * p > n)
                    break;
            }
            if (n != 1) {
                lcm[n] = Math.max(lcm[n], 1);
            }
        }

        for (int i = 0; i < N; i++) {
            int n = a[i];
            long pls = 1;
            for (int p : prime) {
                int cnt = 0;
                // System.out.println(lcm[p]);
                if (lcm[p] > 0) {
                    while (n % p == 0) {
                        n /= p;
                        cnt++;
                    }
                    pls *= pow(p, lcm[p] - cnt);
                    pls %= MOD;
                }
            }
            ans += pls;
            ans %= MOD;
            // System.out.println(ans);
        }
        // System.out.println(Arrays.toString(prime.toArray()));
        System.out.println(ans);
    }

    private static List<Integer> primeList(int max) {
        final List<Integer> prime = new ArrayList<Integer>();
        prime.add(2);
        for (int i = 3; i <= max; i += 2) {
            for (int wari : prime) {
                if (i % wari == 0)
                    break;
                if (wari * wari > i) {
                    prime.add(i);
                    break;
                }
            }
        }
        return prime;
    }

    public static long pow(long x, long n) {
        long sum = 1;
        while (n > 0) {
            if ((n & 1) == 1) {
                sum = sum * x % MOD;
            }
            x = x * x % MOD;
            n >>= 1;
        }
        return sum;
    }
}