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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.math.*;
import java.util.*;
import java.util.function.*;
import java.util.stream.*;

public class Main{
	Scanner s=new Scanner(System.in);

	long mod=1000000007;
	BigInteger MOD=BigInteger.valueOf(1000000007);

	public static BigInteger lcm(BigInteger a,BigInteger b){
		return a.divide(a.gcd(b)).multiply(b);
	}

	void solve(){
		int n=gInt();
		BigInteger[] a=INTS(n).mapToObj(BigInteger::valueOf).toArray(BigInteger[]::new);
		ArrayDeque<BigInteger>q=new ArrayDeque<>();
		for(BigInteger i:a)
			q.add(i);

		while(q.size()>=2){
			BigInteger i=q.poll(),j=q.poll();
			q.add(lcm(i,j));
		}
		BigInteger lcm=q.peek();

		long r=0;
		for(BigInteger o:a){
			r+=lcm.divide(o).mod(MOD).intValue();
			r%=mod;
		}
		System.out.println(r);
	}

	public static void main(String[] args){
		new Main().solve();
	}

	int gInt(){
		return Integer.parseInt(s.next());
	}
	long gLong(){
		return Long.parseLong(s.next());
	}
	double gDouble(){
		return Double.parseDouble(s.next());
	}

	SupplyingIterator<Integer> ints(int n){
		return new SupplyingIterator<>(n,this::gInt);
	}
	SupplyingIterator<Long> longs(int n){
		return new SupplyingIterator<>(n,this::gLong);
	}
	SupplyingIterator<Double> doubles(int n){
		return new SupplyingIterator<>(n,this::gDouble);
	}
	SupplyingIterator<String> strs(int n){
		return new SupplyingIterator<>(n,s::next);
	}

	Range rep(int i){
		return Range.rep(i);
	}
	Range rep(int f,int t,int d){
		return Range.rep(f,t,d);
	}
	Range rep(int f,int t){
		return rep(f,t,1);
	}
	Range rrep(int f,int t){
		return rep(t,f,-1);
	}

	IntStream REP(int v){
		return IntStream.range(0,v);
	}
	IntStream REP(int l,int r){
		return IntStream.rangeClosed(l,r);
	}

	IntStream INTS(int n){
		return IntStream.generate(this::gInt).limit(n);
	}
	LongStream LONGS(int n){
		return LongStream.generate(this::gLong).limit(n);
	}
	Stream<String> STRS(int n){
		return Stream.generate(s::next).limit(n);
	}

}

class SupplyingIterator<T> implements Iterable<T>,Iterator<T>{
	int			t;
	Supplier<T>	supplier;

	SupplyingIterator(int t,Supplier<T> supplier){
		this.t=t;
		this.supplier=supplier;
	}

	@Override
	public Iterator<T> iterator(){
		return this;
	}

	@Override
	public boolean hasNext(){
		return t>0;
	}

	@Override
	public T next(){
		--t;
		return supplier.get();
	}

}

class Range implements Iterable<Integer>,PrimitiveIterator.OfInt{
	int to,cur,d;

	Range(int from,int to,int d){
		this.cur=from-d;
		this.to=to;
		this.d=d;
	}

	Range(int n){
		this(0,n-1,1);
	}

	@Override
	public Iterator<Integer> iterator(){
		return this;
	}

	@Override
	public boolean hasNext(){
		return cur+d==to||(cur!=to&&(cur<to==cur+d<to));
	}

	@Override
	public int nextInt(){
		return cur+=d;
	}
	static Range rep(int i){
		return new Range(i);
	}
	static Range rep(int f,int t,int d){
		return new Range(f,t,d);
	}
	static Range rep(int f,int t){
		return rep(f,t,1);
	}
	static Range rrep(int f,int t){
		return rep(f,t,-1);
	}
}