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: Wrong Answer)

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#define max(x,y)(x>y?x:y)
#define min(x,y)(x>y?y:x)
typedef long long int ll;
typedef unsigned long long int ull;
int asc_cmp(const void* X,const void* Y){int x=*(int*)X,y=*(int*)Y;if(x<y)return -1;else if(x==y)return 0;else return 1;}
int desc_cmp(const void* X,const void* Y){int x=*(int*)X,y=*(int*)Y;if(x<y)return 1;else if(x==y)return 0;else return -1;}
ll gcd(ll x,ll y){if(!y)return x;return gcd(y,x%y);}
ll lcm(ll x,ll y){return x/gcd(x,y)*y;}
int main(){
    int n;scanf("%d",&n);
    static ll a[200000];for(int i=0;i<n;i++)scanf("%lld",&a[i]);
    ll LCM=a[0];
    for(int i=0;i<n-1;i++){
        for(int j=i+1;j<n;j++){
            LCM=lcm(LCM,a[j]);
        }
    }

    ll ans=0;
    for(int i=0;i<n;i++){
        ans+=LCM/a[i];
        ans%=1000000007;
    }
    printf("%lld\n",ans);
}