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: Accepted)


import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
import java.util.concurrent.atomic.AtomicInteger;
import java.util.stream.Collectors;

public class Main {
    private static final long MOD = (long) 1e9 + 7;

    public static void main(final String[] args) {
        final Scanner scanner = new Scanner(System.in);
        final int n = scanner.nextInt();

        final int[] array = new int[n];
        final Map<Integer, Integer> primeCounter = new HashMap<>();
        for (int i = 0; i < n; i++) {
            final int a = scanner.nextInt();
            array[i] = a;
            final Map<Integer, Integer> primeFactorization = primeFactorization(a);
            for (final int key : primeFactorization.keySet()) {
                primeCounter.put(key, Math.max(primeFactorization.get(key), primeCounter.getOrDefault(key, 0)));
            }
        }

        long lcm = 1;
        for (final Map.Entry<Integer, Integer> primeAndCount : primeCounter.entrySet()) {
            lcm *= bigPow(primeAndCount.getKey(), primeAndCount.getValue(), MOD);
            lcm %= MOD;
        }

        long answer = 0;
        for (final long i : array) {
            answer += lcm * modInv(i, MOD);
            answer %= MOD;
        }

        System.out.println(answer);
    }

    /**
     * 素数modを法としてaの逆元を計算する
     * modInvと一緒に使う
     *
     * @param a   逆元を計算したい値
     * @param n   a - 2
     * @param mod 法となる素数
     * @return modを法としたaの逆元
     */
    private static long modPow(long a, long n, final long mod) {
        long res = 1;
        while (n > 0) {
            if ((n & 1) != 0) {
                res = res * a % mod;
            }
            a = a * a % mod;
            n >>= 1;
        }
        return res;
    }

    /**
     * 素数modを法としたaの逆元を計算する
     * modPowと一緒に使う
     *
     * @param a   逆元を計算したい値
     * @param mod 法となる素数
     * @return modを法としたaの逆元
     */
    private static long modInv(final long a, final long mod) {
        return modPow(a, mod - 2, mod);
    }

    private static long bigPow(final long number, final long count, final long mod) {
        long returnValue = 1;
        for (long i = 0; i < count; i++) {
            returnValue *= number;
            returnValue %= mod;
        }
        return returnValue;
    }

    private static Map<Integer, Integer> primeFactorization(long n) {
        final double sqrt = Math.sqrt(n);
        final Map<Integer, AtomicInteger> countMap = new HashMap<>();
        for (int i = 2; i <= sqrt; i++) {
            if (n % i == 0) {
                countMap.computeIfAbsent(i, v -> new AtomicInteger()).incrementAndGet();
                n /= i;
                i--;
            }
        }

        if (n != 1) {
            countMap.computeIfAbsent((int) n, v -> new AtomicInteger()).incrementAndGet();
        }

        return countMap.entrySet().stream()
                .collect(Collectors.toMap(Map.Entry::getKey, entrySet -> entrySet.getValue().get()));
    }
}