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 <limits.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <unistd.h>
#include <math.h>

#define P 1000000007

int comp(const void *a, const void *b){return *(int*)a-*(int*)b;}
int compw(const void *a, const void *b){return (*(int*)a>*(int*)b)-(*(int*)a<*(int*)b);}

uint32_t nextpint(void){ char c=getchar(); while(c<'0'||'9'<c) c=getchar(); uint_fast32_t x=0; while('0'<=c && c<='9'){ x=x*10+c-'0'; c=getchar(); } return x; }

int gcd2(int a, int b){ int t; if(a>b) t=a, a=b, b=t; t = b % a; while(t!=0){ b = a; a = t; t = b % a; } return a; }

int lcm[1000001];

void p(u_int a){
	int g, i, j;
	i=2; g=0; while(a%i==0){a/=i; g++;} if(g>lcm[i]){lcm[i]=g;}
	i=3; g=0; while(a%i==0){a/=i; g++;} if(g>lcm[i]){lcm[i]=g;}
	i=5; g=0; while(a%i==0){a/=i; g++;} if(g>lcm[i]){lcm[i]=g;}
	if(a==1) return;
	int l=sqrt(a);
	for(j=6; j<l; j+=6){
		i=j+1; g=0; while(a%i==0){a/=i; g++;} if(g>lcm[i]){lcm[i]=g;}
		i=j+5; g=0; while(a%i==0){a/=i; g++;} if(g>lcm[i]){lcm[i]=g;}
	}
	if(a!=1){
		if(lcm[a]==0) lcm[a]=1;
	}
}

int r[1000001];
int a[10000];
int main(int argc, char *argv[], char *envp[]){
	r[1]=1;
	for(int i=2; i<=1000000; i++){
		r[i]=P-((long)P/i*r[P%i])%P;
	}
	int n=nextpint();
	for(int i=0; i<n; i++){
		a[i]=nextpint();
		p(a[i]);
	}
	unsigned long f=1;
	for(int i=2; i<=1000000; i++){
		for(int j=lcm[i]-1; j>=0; j--){
			f=f*i%P;
		}
	}
	unsigned long ans=0;
	for(int i=0; i<n; i++){
		ans = (ans + f * r[a[i]]) %P;
	}
	printf("%ld\n", ans);
}