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).

Examples

Input

3
2 3 4


Output

13


Input

5
12 12 12 12 12


Output

5


Input

3
1000000 999999 999998


Output

996989508
#include <bits/stdc++.h>
using namespace std;

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

using ll = long long;

const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
const ll mod = 1000000007LL;

ll pow(ll a, ll n, ll m) {
	ll ret = 1LL;
	for (; n > 0LL; n >>= 1LL, a = a * a % m) {
		if (n % 2LL == 1LL) {
			ret = ret * a % m;
		}
	}
	return ret;
}

int main() {
  ll N;
  cin>>N;
  vector<ll> A(N);
  for(ll i = 0LL; i < N; i++) cin>>A[i];
  map<ll,ll> m;
  for(auto x:A){
    ll y = x;
    map<ll,ll> mx;
    for(ll i = 2LL; i*i<=x;i++){
      while(y%i==0LL){
        y/=i;
        mx[i]++;
      }
    }
    if(y!=1LL) mx[y]++;
    for(auto t:mx) chmax(m[t.first],t.second);
  }
  ll all=1LL;
  for(auto x:m){
    all*=pow(x.first,x.second,mod);
    all%=mod;
  }
  ll ans = 0LL;
  for(auto x:A){
    ans+=all*pow(x,mod-2,mod);
    ans%=mod;
  }
  cout<<ans<<endl;
}