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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
typedef long long ll;
typedef pair<int, int> P;

ll mod = 1e9 + 7;

ll mpow(ll x, ll n){ //x^n(mod mod) ←普通にpow(x,n)では溢れてしまうため，随時mod計算
    ll ans = 1;
    while(n != 0){
        if(n&1) ans = ans*x % mod;
        x = x*x % mod;
        n = n >> 1;
    }
    return ans;
}

int main() {
    int N; cin >> N;
    vector<ll> A(N);
    rep(i, N){
        cin >> A[i];
    }
    vector<ll> a(N);
    rep(i, N){
        a[i] = A[i];
    }
    map<int, int> l;
    vector<int> check(1001);
    for (int i = 2; i < 1001; i++){
        if(check[i] != 0)continue;
        for (int j = i*i; j < 1001; j+=i){
            check[j] = 1;
        }
        int cnt = 0;
        rep(k, N){
            int cntnow = 0;
            while(a[k] % i == 0){
                a[k] /= i;
                cntnow++;
            }
            cnt = max(cnt, cntnow);
        }
        l[i] = cnt;
    }

    ll lc = 1;
    for (auto p : l){
        int key = p.first;
        int value = p.second;
        rep(i, value){
            lc *= key;
            lc %= mod;
        }
    }

    ll ans = 0;
    rep(i, N){
        ans += lc * mpow(A[i], mod-2);
        ans %= mod;
    }

    cout << ans << endl;

    return 0;
}