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<iostream>
#include<vector>
#include<algorithm>
#include<map>
#include<cstring>
#include<math.h>
#include<bitset>
#include<queue>
#include<set>
#include<iomanip>
#include<math.h>
#include<assert.h>
#include<string>
#include<cstdlib>
#include<cctype>
// #include<bits/stdc++.h>
using namespace std;
typedef long long ll;
constexpr ll INF = 1LL << 40;
const double INFD = 1e100;
const ll mod = 1000000007;
const double PI = 3.1415926535897932384626433832795028841971;
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
// ----------------------------------------------------------------------------
ll gcd(ll a,ll b){
  if(b == 0) return a;
  return gcd(b,a%b);
}

int main(){
  int N;
  cin >> N;
  vector<ll> A(N);
  for(int i=0; i<N; i++){
    cin >> A[i];
  }
  ll lc=A[0];
  for(int i=1; i<N; i++){
    lc = lc/gcd(lc,A[i])*A[i];
  }
  ll sum=0;
  for(int i=0; i<N; i++){
    sum += (lc/A[i])%mod;
    sum %= mod;
  }
  cout << sum << endl;
  return 0;
}