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++ (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;
const int N=1e6+100;
const int mod=1e9+7;
typedef long long ll;
ll num[N],a[N];

ll qpow(ll x,ll n)
{
    ll sum=1;
    while(n){
        if(n&1) sum=sum*x%mod;
        x=x*x%mod;
        n>>=1;
    }
    return sum;
}
ll maxx=0;
int main()
{
    int n;scanf("%d",&n);
    for(int i=1;i<=n;i++) {
        scanf("%lld",&a[i]);
        maxx=max(maxx,a[i]);
    }
  //  cout<<maxx<<endl;
    for(int i=1;i<=n;i++){
            int z=a[i];
            int zz=a[i];
        for(int j=2;j<=sqrt(zz);j++){
                ll cnt=0;
            while(z%j==0) {
                cnt++;
                z/=j;
            }
            num[j]=max(num[j],cnt);
//            cout<<j<<" "<<num[j]<<endl;
        }
      if(z!=1)  num[z]=max(1ll,num[z]);
    }
//    for(int i=1;i<=maxx;i++)
//         cout<<i<<" "<<num[i]<<endl;
    ll sum=1;
    for(int i=2;i<=maxx;i++){
        sum=(sum%mod*qpow(i,num[i])%mod)%mod;
    }
  //  cout<<sum<<endl;
    ll ans=0;
    for(int i=1;i<=n;i++){
        ans=(ans+sum*qpow(a[i],mod-2))%mod;
    }
    printf("%lld\n",ans);
    return 0;
}