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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;

const int Nmax = 1e4 + 1;
const int Vmax = 1e6 + 5;
const int oo = LONG_MAX;
const int Mod = 1e9 + 7;

int n;
int a[Nmax], maxPow[Vmax];

void Factori(int x)
{
    int p = 0;

    while(x%2 == 0)
    {
        p++;
        x/=2;
    }

    maxPow[2] = max(maxPow[2], p);

    int d = 3;

    while(x > 1 && d * d <= x)
    {
        p = 0;
        while(x%d == 0)
        {
            p++;
            x/=d;
        }

        maxPow[d] = max(maxPow[d], p);
        d += 2;
    }

    if(x > 1) maxPow[x] = max(maxPow[x], 1);
}

long long Lgput(long long x, int y)
{
    long long p = 1;

    while(y > 0)
    {
        if(y%2 == 1) p = p * x;
        x = x * x;
        y=y/2;
    }

    return p;
}

int main()
{ /**
    #ifndef ONLINE_JUDGE

        ifstream cin("test.in");

    #endif // ONLINE_JUDGE
*/
    cin >> n;

    for(int i = 1; i <= n; i++)
    {
        cin >> a[i];
        Factori(a[i]);
    }


    long long lcm = 1;

    for(int i = 2; i < Vmax; i++)
        if(maxPow[i])lcm *= Lgput(i, maxPow[i]);

    long long sum = 0;
    for(int i = 1; i <= n; i++)
        sum = (sum + lcm/a[i]) % Mod;

    cout << sum;
    return 0;
}