/**
	* Author    : Arif Ahmad
	* Date      :
	* Algo      :
	* Difficulty:
*/
 
#include <bits/stdc++.h>
using namespace std;
 
#define INF_MAX 	2147483647
#define INF_MIN 	-2147483648
#define INF 		(1 << 30)
#define EPS			1e-9
#define PI 			acos(-1.0)
#define N 		    112345
#define MOD			10000000007
#define sz(x) 		(int)(x).size()
#define all(x) 		(x).begin(), (x).end()
#define pb 			push_back
#define mp			make_pair
#define ms(x, a) 	memset((x), (a), sizeof(x))
#define F           first
#define S           second
#define rep(i,a,b)  for(int i=(a); i<(b); ++i)
#define repC(i,x) 	for(size_t i=0; i<x.size(); ++i)
#define repIT(it,c) for(__typeof((c).begin()) it=(c).begin(); it!=(c).end(); ++it)
#define nn          '\n'
#define trace(x)    cerr << #x << " = " << x << '\n';
#define trace2(x,y)	cerr << #x << "=" << x << "\t" << #y << "=" << y << '\n';
 
typedef long long 		LL;
typedef pair<int,int> 	pii;
typedef vector<int> 	vi;
typedef vector<string> 	vs;
typedef vector<char>	vc;
typedef vector<bool>    vb;
typedef vector< pii >   vii;
typedef map<string,int> msi;
typedef map<int,int>	mii;
typedef map<char,int>   mci;
typedef map<int,string>	mis;
 
template<class T> T Abs(T x) {return x>0 ? x : -x;}
template<class T> T Max(T a, T b) {return a>b ? a : b;}
template<class T> T Min(T a, T b) {return a<b ? a : b;}
template<class T> T gcd(T a, T b) {return (b ? gcd(b,a%b) : a);}
bool isVowel(char ch){ch=tolower(ch);return(ch=='a' || ch=='e' || ch=='i' || ch=='o' || ch=='u');}
 
//int dx[4] = {-1, 0, 0, 1};
//int dy[4] = {0, -1, 1, 0};
//int dx[8] = {-1, -1, -1, 0, 0, 1, 1, 1};
//int dy[8] = {-1, 0, 1, -1, 1, -1, 0, 1};
 
vi factors;
 
void prime_factorize(int n) 
{
	factors.clear();
	factors.resize(1000, 0);
 
	for(int i = 2; i * i <= n; i++) {
		while(n % i == 0) {
			factors[i]++;
			n /= i;
		}
	}
	if(n > 1) {
		factors[n]++;
	}
}
 
int main()
{
    ios_base :: sync_with_stdio(0);
    cin.tie(0);
 
 
	int i, j, k, n, m, tc;
 
    cin >> tc;
	while(tc--) {
		cin >> n;
 
		prime_factorize(n);
 
		int ans = 1;
		for(i = 2; i < 1000; i++) {
			if(factors[i] > 0 && factors[i] % 2 == 1) {
				ans *= i;	
			}
		}
 
		cout << ans << nn;
	}
 
	return 0;
}