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 Java7 (OpenJDK 1.7.0) (judge verdict: Wrong Answer)

import java.util.*;
public class Main {
  	private static long MOD = 1000000007;
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);

		// 整数の入力
		int a = sc.nextInt();
      	long aaa[] = new long[a];
        ArrayList<Long> list = new ArrayList<Long>();
		for (int i=0; i < a; i++) {
			list.add(sc.nextLong());
		}
		long max = getLcm(list) % MOD;
		long ans = 0;
		for (int i=0; i < list.size(); i++) {
			ans += (max / list.get(i)) % MOD;
		}
		System.out.println(ans% MOD);

	}
	public static long getLcm(ArrayList<Long> list){
        long valueLCM = 0;

        for(int index=0; index<list.size()-1; index++) {
            if( index==0 ) {
                valueLCM = calcLCM( list.get(index), list.get(index+1) );
            } else {
                valueLCM = calcLCM( valueLCM,        list.get(index+1) );
            }
        }
        return valueLCM;
    }

    private static long calcLCM(long val1, long val2) {
        long maxValue = Math.max(val1, val2);
        long minValue = Math.min(val1, val2);
        long val3    = maxValue * minValue % MOD;

        if(minValue==0) return maxValue;

        long temp;
        while( (temp=maxValue%minValue)!=0 ) {
            maxValue=minValue;
            minValue=temp;
        }

        return (long)(val3/minValue % MOD) ;
    }
}