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>
int main()
{
  long int N,i,j,m;
  long long int k=0;
  int number;
  int digit1,digit2;
  int initial1,initial2;
  scanf("%ld",&N);
  for(i=1;i<N+1;i++){
    digit1=0;
    number=i;
    initial1=i;
    while(number!=0){
    number = number/ 10;
    digit1++;
  }
    digit1=digit1-1;
    if(digit1!=0){
    for(m=0;m<digit1;m++){
      initial1=initial1/10;
    }
    }
    
    for(j=1;j<N+1;j++){
     digit2=0;
    number=j;
      initial2=j;
    while(number!=0){
    number = number / 10;
    digit2++;
  }
    digit2=digit2-1;
    if(digit2!=0){
    for(m=0;m<digit2;m++){
      initial2=initial2/10;
    }
    }
      if((i%10==initial2)&&(j%10==initial1)){
        k++;
      }
    }
  }
  printf("%lld",k);
  return 0;
}