// Slower C++ solution:
#include <bits/stdc++.h>

#define mst(a,b) memset(a,b,sizeof(a))

#define lowbit(x) ((x)&(-x))

#define X first

#define Y second

using namespace std;

typedef long long LL;

typedef long long ll;

typedef unsigned long long ull;

typedef pair<int, int> pii;

typedef pair<ll, ll> pll;

//const int mod = 1e9+7;

const int inf = 0x3f3f3f3f;

const ll INF = 0x3f3f3f3f3f3f3f3f;

const int maxn = 100000+10;

const double eps = 1e-9;

//const int maxm = 2510000+10;

bool is_prime[maxn];

int prime[maxn],id[maxn],tot;

ull seed=131;

ull f[maxn];

map<ull,int> cnt;

pair<ull,ull> a[maxn];

void init(){

    is_prime[1]=true;

    f[0]=1;

    for (int i=2;i*i<=100000;i++)

        if (!is_prime[i])

            for (int j=i*i;j<=100000;j+=i) is_prime[j]=true;

    for (int i=1;i<=100000;i++){

        if (!is_prime[i]) {

            prime[tot]=i;

            id[i]=tot++;

        } else id[i]=-1;

        f[i]=f[i-1]*seed;

    }

}

int main() {

#ifdef local

    freopen("in.txt", "r", stdin);

//    freopen("out.txt", "w", stdout);

#endif

    ios::sync_with_stdio(0);

    cin.tie(0);cout.tie(0);

    init();

    int n; cin>>n;

    int ans=0;

    for (int i=0;i<n;i++){

        ll x; cin>>x;

        ull H=0,P=0;

        for (int j=0;j<350 && 1ll*prime[j]*prime[j]<=x;j++){

            if (x%prime[j]==0){

                int c=0;

                while (x%prime[j]==0){

                    c=(c+1)%3;

                    x/=prime[j];

                }

                H+=f[id[prime[j]]]*c;

                P+=f[id[prime[j]]]*((3-c)%3);

            }

        }

        if (x>1){

            if (x<=100000 && !is_prime[x]){

                H+=f[id[x]];

                P+=f[id[x]]*2;

            } else if (x>100000) {

                ll tmp=sqrt(x)+eps;

                if (tmp*tmp==x && !is_prime[tmp]){

                    H+=f[id[tmp]]*2;

                    P+=f[id[tmp]];

                } else {

                    ans++;

                    continue;

                }

            }

        }

        a[i].X=H; a[i].Y=P;

        cnt[H]++;

    }

    if (cnt.count(0)) {

        ans++;

        cnt[0]=0;

    }

    for (int i=0;i<n;i++){

        ull H=a[i].X,P=a[i].Y;

        ans+=max(cnt[H],cnt[P]);

        cnt[H]=0; cnt[P]=0;

    }

    cout<<ans<<"\n";

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 2.3994101088356343x):
#include <bits/stdc++.h>

using namespace std;

#define int long long

const int maxN = 2e3 + 13, maxK = 1e5 + 13;

int n, b[maxK], a[maxK], ans;

map<int, int> val;

map<int, bool> mark;

int32_t main() {

    ios::sync_with_stdio(false);

    cin.tie(0);

    cout.tie(0);

    cin >> n;

    for (int i = 0; i < n; i++) {

	cin >> a[i];

	int x = a[i];

	a[i] = b[i] = 1;

	for (int j = 2; j < maxN; j++) {

	    if(x % j == 0){

		int cnt = 0;

		while (x % j == 0) {

		    x /= j;

		    cnt++;

		}

		if (cnt % 3 == 1) {

		    a[i] *= j;

		    b[i] *= j * j;

		}

		else if (cnt % 3 == 2){

		    a[i] *= j * j;

		    b[i] *= j;

		}

	    }

	}

	if(x != 1){

	    int y = ceil(sqrt(x));

	    a[i] *= x;

	    if (y * y == x)

		b[i] *= y;

	    else

		b[i] *= x * x;

	}

	val[a[i]]++;

    }

    for (int i = 0; i < n; i++)

	if (!mark[a[i]]) {

	    mark[a[i]] = mark[b[i]] = 1;

	    if (a[i] == 1)

		ans++;

	    else

		ans += max(val[a[i]], val[b[i]]);

	}



    cout << ans << endl;

}