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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;

#define rep(i, n) for(int i = 0; i < (n); ++i)

typedef long long ll;
typedef pair<int, int> P;

const int MAX = 1e5 + 5;
const int INF = 1001001001;
const int MOD = 1e9 + 7;

map<int, int> prime_factor(int n) {
    map<int, int> res;
    for (int i = 2; i * i <= n; ++i) {
        while (n % i == 0) {
            ++res[i];
            n /= i;
        }
    }
    if (n != 1) res[n] = 1;
    return res;
}

ll mod_pow(ll x, ll y) {
    ll res = 1;
    while (y > 0) {
        if (y & 1) res = res * x % MOD;
        x = x * x % MOD;
        y >>= 1;
    }
    return res;
}

int main(){
    int N;
    cin >> N;

    vector<int> A(N);
    rep(i, N) cin >> A[i];

    vector<map<int, int>> nums(N);
    rep(i, N) nums[i] = prime_factor(A[i]);

    map<int, int> lcm;
    for (auto &a : nums) {
        for (auto &p : a) {
            lcm[p.first] = max(lcm[p.first], p.second);
        }
    }

    ll ans = 0;
    for (auto &a : nums) {
        ll res = 1;
        for (auto &p : lcm) {
            res *= mod_pow(p.first, p.second - a[p.first]);
            res %= MOD;
        }
        ans += res;
        ans %= MOD;
    }

    cout << ans << endl;
}