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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/trie_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> pbds;
typedef trie<string, null_type, trie_string_access_traits<>, pat_trie_tag, trie_prefix_search_node_update> pref_trie;
typedef unsigned long long BITMASK; typedef long long ll; typedef long double ld;
#define has_bit(bit_mask, x) ((bit_mask) & (1Uint << (x)))
#define turn_on_bit(bit_mask, x) (bit_mask |= (1Uint << (x)))
#define turn_off_bit(bit_mask, x) (bit_mask &= (~(1Uint << (x))))
#define smallest_on_bit(bit_mask) (__builtin_ctzint((bit_mask) & (-(bit_mask))))
#define CLOCK_START clock_t chrono_clk_beg = clock()
#define CLOCK_END clock_t chrono_clk_end = clock(); cerr << (double(chrono_clk_end - chrono_clk_beg) / CLOCKS_PER_SEC) << " sec"
#define bug(args ...) cerr << __LINE__ << ">> ", err(new istringstream(string(#args)), args), cerr << '\n'
#define all(x) x.begin(), x.end()
#define mapfind(mp, x) (mp.find(x) != mp.end())
#define removeDuplicates(vec) sort(vec.begin(), vec.end()); vec.erase(unique(vec.begin(), vec.end()), vec.end())
#define int ll
#define double ld
void err(istringstream *iss) {} template<typename T, typename ... Args> void err(istringstream *iss, const T &_val, const Args & ... args) { string _name; *iss >> _name; if (_name.back()==',') _name.pop_back(); cerr << _name << " = " << _val << "; ", err(iss, args ...); }
#define MOD 1000000007

signed main()
{
    ios_base::sync_with_stdio(false); cin.tie(NULL);
    int n;
    cin >> n;
    vector<int> arr(n);
    for (int i = 0; i < n; ++i) cin >> arr[i];
    int lcm = arr[0];
    for (int i = 1; i < n; ++i)
        lcm = ((arr[i] * lcm) / __gcd(arr[i], lcm));
    int ans = 0;
    for (int i = 0; i < n; ++i)
        ans = (ans + (lcm / arr[i])) % MOD;
    cout << ans << '\n';
    return 0;
}