将 x 分解质因子后判断奇偶即可。
#include<bits/stdc++.h>
#define int long long
#define double long double
#define x first
#define y second
using namespace std;
typedef long long LL;
typedef long long ll;
typedef pair<int, int> PII;
const int N = 3e5 + 10;
const int M = 1e3 + 10;
int mod = 1e9 + 7;
int a[N];
void solve() {
int n;
cin >> n;
int sum = 0;
for (int i = 2; i * i <= n; i++) {
while (n % i == 0) n /= i, sum++;
}
if (n > 1) sum++;
if (sum % 2) cout << "kou\n";
else cout << "yukari\n";
}
signed main() {
ios::sync_with_stdio(false);
cin.tie(0), cout.tie(0);
int _;
_ = 1;
cin >> _;
while (_--) {
solve();
}
}

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