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 <vector>
#include <iostream>
#include <algorithm>
#include <unordered_map>
using namespace std;
const int mod = 1000000007;
int binpow(int a, int b) {
	int ans = 1;
	while (b > 0) {
		if (b & 1) ans = 1LL * ans * a % mod;
		a = 1LL * a * a % mod;
		b >>= 1;
	}
	return ans;
}
int main() {
	cin.tie(0);
	ios_base::sync_with_stdio(false);
	int N;
	cin >> N;
	vector<int> A(N);
	int invsum = 0;
	for (int i = 0; i < N; ++i) {
		cin >> A[i];
		invsum = (invsum + binpow(A[i], mod - 2)) % mod;
	}
	int mx = *max_element(A.begin(), A.end());
	int b = 0;
	while (b * b < mx) ++b;
	vector<bool> prime(b + 1, true);
	prime[0] = false; prime[1] = false;
	for (int i = 2; i <= b; ++i) {
		if (!prime[i]) continue;
		for (int j = i * 2; j <= b; j += i) {
			prime[j] = false;
		}
	}
	vector<int> p;
	for (int i = 2; i <= b; ++i) {
		if (prime[i]) {
			p.push_back(i);
		}
	}
	unordered_map<int, int> d;
	for (int i = 0; i < N; ++i) {
		for (int j = 0; j < p.size() && p[j] * p[j] <= A[i]; ++j) {
			if (A[i] % p[j] == 0) {
				int mul = 1;
				while (A[i] % p[j] == 0) A[i] /= p[j], mul *= p[j];
				d[p[j]] = max(d[p[j]], mul);
			}
		}
		if (A[i] > 1) {
			d[A[i]] = max(A[i], A[i]);
		}
	}
	int pmul = 1;
	for (pair<int, int> i : d) {
		pmul = 1LL * pmul * i.second % mod;
	}
	int ans = 1LL * invsum * pmul % mod;
	cout << ans << endl;
	return 0;
}