Problem:
Score : 1100 points

Problem Statement

Snuke got positive integers s_1,...,s_N from his mother, as a birthday present. There may be duplicate elements.

He will circle some of these N integers. Since he dislikes cubic numbers, he wants to ensure that if both s_i and s_j (i ≠ j) are circled, the product s_is_j is not cubic. For example, when s_1=1,s_2=1,s_3=2,s_4=4, it is not possible to circle both s_1 and s_2 at the same time. It is not possible to circle both s_3 and s_4 at the same time, either.

Find the maximum number of integers that Snuke can circle.

Constraints

1 ≦ N ≦ 10^5

1 ≦ s_i ≦ 10^{10}

All input values are integers.

Input

The input is given from Standard Input in the following format:

N
s_1
:
s_N

Output

Print the maximum number of integers that Snuke can circle.

Sample Input 1

8
1
2
3
4
5
6
7
8

Sample Output 1

6

Snuke can circle 1,2,3,5,6,7.

Sample Input 2

6
2
4
8
16
32
64

Sample Output 2

3

Sample Input 3

10
1
10
100
1000000007
10000000000
1000000009
999999999
999
999
999

Sample Output 3

9

Solution in C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Numerics;

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

public class Program
{
    private int N;
    private long[] S;

    private List<int> Primes = new List<int>();

    public void Solve()
    {
        var sc = new Scanner();
        N = sc.NextInt();
        S = new long[N];
        for (int i = 0; i < N; i++)
        {
            S[i] = sc.NextLong();
        }

        // S[i]の各素因数の次数をmod3した数の個数
        var c = new HashMap<long, int>();

        // S[i]の各素因数の次数をmod3した数 normに p[norm]をかけると立方数になる
        var p = new HashMap<long, long>();

        // sqrt(10^10) = 10^5までの素数表を作成
        MathEx.Sieve(100000, Primes);

        var dict = new Dictionary<long, long>();

        foreach (int i in Primes)
        {
            dict[(long) i * i] = i;
        }

        bool flag = false;
        long ans = 0;
        foreach (long l in S)
        {
            long tmpL = l;
            bool overflowFlag = false;
            BigInteger pair = 1;
            long norm = 1;
            foreach (int i in Primes)
            {
                long pow3 = (long) i * i * i;
                if (pow3 > tmpL)
                {
                    break;
                }

                while (tmpL % pow3 == 0)
                {
                    tmpL /= pow3;
                }

                long pow2 = (long) i * i;

                if (tmpL % pow2 == 0)
                {
                    norm *= pow2;
                    pair *= norm;
                    tmpL /= pow2;
                }
                else if (tmpL % i == 0)
                {
                    norm *= i;
                    pair *= pow2;
                    tmpL /= i;
                }
            }

           
            long o;
            if (dict.TryGetValue(tmpL,out o))
            {
                // tmpLが平方数
                norm *= tmpL;
                pair *= o;
            }
            else
            {
                // tmpLが素数 or 2素数の積
                norm *= tmpL;
                pair *= tmpL;
                pair *= tmpL;
            }

            if (norm == 1)
            {
                flag = true;
            }
            else if (pair > 10000000000)
            {
                ans++;
            }
            else
            {
                c[norm]++;
                p[norm] = (long)pair;
            }
        }


        var keys = c.Keys.ToArray();

        foreach (long key in keys)
        {
            ans += Math.Max(c[key], c[p[key]]);
            c[key] = 0;
            c[p[key]] = 0;
        }

        if (flag) ans++;
        Console.WriteLine(ans);
    }

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

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
}