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: Accepted)

#include"bits/stdc++.h"
using namespace std;
typedef long long ll;
const ll MOD = (ll)(1e9+7);
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define rep(i, n) for(int (i)=0; (i)<(int)(n); (i)++)
#ifdef LOCAL
#define debug(x) cerr << #x << ": " << x << endl
#else
#define debug(x)
#endif
int dx[4]={ 1,0,-1,0 };
int dy[4]={ 0,1,0,-1 };

//累乗 O(log N)
long long power(long long x, long long N){
	if(N == 0) return 1LL;
	if(N == 1) return x;
	long long tmp = power(x, N/2);
	if(N%2 == 0) return tmp * tmp % MOD;
	else return tmp * tmp % MOD * x % MOD;
}

//逆元 O(log x)
long long inverse(long long x){
	return power(x, MOD-2);
}

int N;
ll A[10000];
ll so[1000000] ={};
signed main(){
	cin >> N;
	rep(i, N) cin >> A[i];

	rep(i, N){
		ll a = A[i];
		for(ll j=2; j<=sqrt(A[i]); j++){
			ll c = 0;
			while(a%j == 0){
				c++;
				a /= j;
			}
			so[j] = max(so[j], c);
		}
		if(a > 1){
			so[a] = max(so[a], 1LL);
		}
	}

	ll sum = 1;
	for(ll i=2; i<1000000; i++){
		rep(j, so[i]){
			sum *= i;
			sum %= MOD;
		}
	}

	ll ans = 0;
	rep(i, N){
		ans += sum * inverse(A[i]);
		ans %= MOD;
	}
	cout << ans << endl;
}