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# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;

public class Program
{
    private int N;
    private int[] A;


    public void Solve()
    {
        var sc = new Scanner();
        N = sc.NextInt();
        A = sc.IntArray();
        var primes = new List<int>();
        MathEx.Sieve(1000, primes);

        ModInt[] B = new ModInt[N];
        for (int i = 0; i < N; i++)
        {
            B[i] = 1;
        }

        foreach (int prime in primes)
        {
            int max = 0;
            int[] cnt = new int[N];
            for (int i = 0; i < N; i++)
            {
                while (A[i] % prime == 0)
                {
                    cnt[i]++;
                    A[i] /= prime;
                }

                max = Math.Max(max, cnt[i]);
            }

            for (int i = 0; i < N; i++)
            {
                B[i] *= ModInt.Pow(prime, max - cnt[i]);
            }
        }

        for (int i = 0; i < N; i++)
        {
            B[i] *= A[i];
        }

        ModInt ans = 0;
        foreach (ModInt i in B)
        {
            ans += i;
        }

        Console.WriteLine(ans);
    }

    public static void Main(string[] args)
    {
        new Program().Solve();
    }
}

#region ModInt

/// <summary>
/// [0,<see cref="Mod"/>) までの値を取るような数
/// </summary>
public struct ModInt
{
    /// <summary>
    /// 剰余を取る値．
    /// </summary>
    public const long Mod = (int) 1e9 + 7;

    /// <summary>
    /// 実際の数値．
    /// </summary>
    public long num;

    /// <summary>
    /// 値が <paramref name="n"/> であるようなインスタンスを構築します．
    /// </summary>
    /// <param name="n">インスタンスが持つ値</param>
    /// <remarks>パフォーマンスの問題上，コンストラクタ内では剰余を取りません．そのため，<paramref name="n"/> ∈ [0,<see cref="Mod"/>) を満たすような <paramref name="n"/> を渡してください．このコンストラクタは O(1) で実行されます．</remarks>
    public ModInt(long n)
    {
        num = n;
    }

    /// <summary>
    /// このインスタンスの数値を文字列に変換します．
    /// </summary>
    /// <returns>[0,<see cref="Mod"/>) の範囲内の整数を 10 進表記したもの．</returns>
    public override string ToString()
    {
        return num.ToString();
    }

    public static ModInt operator +(ModInt l, ModInt r)
    {
        l.num += r.num;
        if (l.num >= Mod) l.num -= Mod;
        return l;
    }

    public static ModInt operator -(ModInt l, ModInt r)
    {
        l.num -= r.num;
        if (l.num < 0) l.num += Mod;
        return l;
    }

    public static ModInt operator *(ModInt l, ModInt r)
    {
        return new ModInt(l.num * r.num % Mod);
    }

    public static implicit operator ModInt(long n)
    {
        n %= Mod;
        if (n < 0) n += Mod;
        return new ModInt(n);
    }

    /// <summary>
    /// 与えられた 2 つの数値からべき剰余を計算します．
    /// </summary>
    /// <param name="v">べき乗の底</param>
    /// <param name="k">べき指数</param>
    /// <returns>繰り返し二乗法により O(N log N) で実行されます．</returns>
    public static ModInt Pow(ModInt v, long k)
    {
        return Pow(v.num, k);
    }

    /// <summary>
    /// 与えられた 2 つの数値からべき剰余を計算します．
    /// </summary>
    /// <param name="v">べき乗の底</param>
    /// <param name="k">べき指数</param>
    /// <returns>繰り返し二乗法により O(N log N) で実行されます．</returns>
    public static ModInt Pow(long v, long k)
    {
        long ret = 1;
        for (k %= Mod - 1; k > 0; k >>= 1, v = v * v % Mod)
            if ((k & 1) == 1)
                ret = ret * v % Mod;
        return new ModInt(ret);
    }

    /// <summary>
    /// 与えられた数の逆元を計算します．
    /// </summary>
    /// <param name="v">逆元を取る対象となる数</param>
    /// <returns>逆元となるような値</returns>
    /// <remarks>法が素数であることを仮定して，フェルマーの小定理に従って逆元を O(log N) で計算します．</remarks>
    public static ModInt Inverse(ModInt v)
    {
        return Pow(v, Mod - 2);
    }
}

#endregion

class HashMap<K, V> : Dictionary<K, V>
{
    new public V this[K i]
    {
        get
        {
            V v;
            return TryGetValue(i, out v) ? v : default(V);
        }
        set { base[i] = value; }
    }
}

#region GCD LCM

/// <summary>
/// 様々な数学的関数の静的メソッドを提供します．
/// </summary>
public static partial class MathEx
{
    /// <summary>
    /// 2 つの整数の最大公約数を求めます．
    /// </summary>
    /// <param name="n">最初の値</param>
    /// <param name="m">2 番目の値</param>
    /// <returns>2 つの整数の最大公約数</returns>
    /// <remarks>ユークリッドの互除法に基づき最悪計算量 O(log N) で実行されます．</remarks>
    public static int GCD(int n, int m)
    {
        return (int) GCD((long) n, m);
    }


    /// <summary>
    /// 2 つの整数の最大公約数を求めます．
    /// </summary>
    /// <param name="n">最初の値</param>
    /// <param name="m">2 番目の値</param>
    /// <returns>2 つの整数の最大公約数</returns>
    /// <remarks>ユークリッドの互除法に基づき最悪計算量 O(log N) で実行されます．</remarks>
    public static long GCD(long n, long m)
    {
        n = Math.Abs(n);
        m = Math.Abs(m);
        while (n != 0)
        {
            m %= n;
            if (m == 0) return n;
            n %= m;
        }

        return m;
    }


    /// <summary>
    /// 2 つの整数の最小公倍数を求めます．
    /// </summary>
    /// <param name="n">最初の値</param>
    /// <param name="m">2 番目の値</param>
    /// <returns>2 つの整数の最小公倍数</returns>
    /// <remarks>最悪計算量 O(log N) で実行されます．</remarks>
    public static long LCM(long n, long m)
    {
        return (n / GCD(n, m)) * m;
    }
}

#endregion

#region PrimeSieve

public static partial class MathEx
{
    /// <summary>
    /// ある値までに素数表を構築します．
    /// </summary>
    /// <param name="max">最大の値</param>
    /// <param name="primes">素数のみを入れた数列が返される</param>
    /// <returns>0 から max までの素数表</returns>
    /// <remarks>エラトステネスの篩に基づき，最悪計算量 O(N loglog N) で実行されます．</remarks>
    public static bool[] Sieve(int max, List<int> primes = null)
    {
        var isPrime = new bool[max + 1];
        for (int i = 2; i < isPrime.Length; i++) isPrime[i] = true;
        for (int i = 2; i * i <= max; i++)
            if (!isPrime[i]) continue;
            else
                for (int j = i * i; j <= max; j += i)
                    isPrime[j] = false;
        if (primes != null)
            for (int i = 0; i <= max; i++)
                if (isPrime[i])
                    primes.Add(i);

        return isPrime;
    }
}

#endregion

class Scanner
{
    public Scanner()
    {
        _pos = 0;
        _line = new string[0];
    }

    const char Separator = ' ';
    private int _pos;
    private string[] _line;

    #region スペース区切りで取得

    public string Next()
    {
        if (_pos >= _line.Length)
        {
            _line = Console.ReadLine().Split(Separator);
            _pos = 0;
        }

        return _line[_pos++];
    }

    public int NextInt()
    {
        return int.Parse(Next());
    }

    public long NextLong()
    {
        return long.Parse(Next());
    }

    public double NextDouble()
    {
        return double.Parse(Next());
    }

    #endregion

    #region 型変換

    private int[] ToIntArray(string[] array)
    {
        var result = new int[array.Length];
        for (int i = 0; i < array.Length; i++)
        {
            result[i] = int.Parse(array[i]);
        }

        return result;
    }

    private long[] ToLongArray(string[] array)
    {
        var result = new long[array.Length];
        for (int i = 0; i < array.Length; i++)
        {
            result[i] = long.Parse(array[i]);
        }

        return result;
    }

    private double[] ToDoubleArray(string[] array)
    {
        var result = new double[array.Length];
        for (int i = 0; i < array.Length; i++)
        {
            result[i] = double.Parse(array[i]);
        }

        return result;
    }

    #endregion

    #region 配列取得

    public string[] Array()
    {
        if (_pos >= _line.Length)
            _line = Console.ReadLine().Split(Separator);

        _pos = _line.Length;
        return _line;
    }

    public int[] IntArray()
    {
        return ToIntArray(Array());
    }

    public long[] LongArray()
    {
        return ToLongArray(Array());
    }

    public double[] DoubleArray()
    {
        return ToDoubleArray(Array());
    }

    #endregion
}