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;

using ll = long long;

constexpr ll MOD = 1000000007;

int N;
vector<int> A(10001);

ll modpow(ll n, ll k) {
    if(k == 0) return 1;
    if(k % 2 == 0) {
        ll tmp = modpow(n, k/2);
        return (tmp * tmp) % MOD;
    }
    return (n * modpow(n, k - 1)) % MOD;
}

void solve(){
    map<int, int> pf;
    for(int i = 1; i <= N; ++i){
        map<int, int> tmp_pf;
        int a = A[i];
        for(int k = 2; k <= a && k * k <= A[i]; ++k){
            if(a % k == 0) {
                tmp_pf[k]++;
                a /= k;
            }
        }
        tmp_pf[a]++;
        for(auto itr = tmp_pf.begin(); itr != tmp_pf.end(); ++itr){
            pf[itr->first] = max(pf[itr->first], itr->second);
        }
    }

    ll lcm = 1;
    for(auto itr = pf.begin(); itr != pf.end(); ++itr){
        lcm *= modpow(itr->first, itr->second);
        lcm %= MOD;
    }

    ll ans = 0;
    for(int i = 1; i <= N; ++i){
        ans += (lcm * modpow(A[i], MOD - 2)) % MOD;
        ans %= MOD;
    }
    cout << ans;
}
int main(){
    cin >> N;
    for(int i = 1; i <= N; ++i){
        cin >> A[i];
    }
    solve();
    return 0;
}