n的约数

#include<bits/stdc++.h>
using namespace std;
const int N = 1e5 + 7;
typedef long long LL;
const LL M = 1000000000000000000;
const LL p[17] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59};
int prime[N]; LL ans, n;
void getPrime() {
    memset(prime, 0, sizeof(prime));
    for (int i = 2; i < N; i++) {
        if (!prime[i]) prime[++prime[0]] = i;
        for (int j = 1; j <= prime[0] && prime[j] * i < N; j++) {
            prime[prime[j] * i] = 1;
            if (i % prime[j] == 0) break;
        }
    }
}
inline LL read() {
#define isDigit(ch) (ch >= '0' && ch <= '9')
    char ch = getchar();
    LL x = 0, f = 1;
    while (!isDigit(ch)) {
        if (ch == '-') f = -1;
        ch = getchar();
    }
    while (isDigit(ch)) {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x * f;
}
void dfs(int ps, LL cur, LL c, int pre) {
    if (ps > 17) return;
    ans = max(ans, c);
    LL now = cur;
    for (int i = 1; i <= pre && n / p[ps] >= now; i++) {
        now *= p[ps];
        dfs(ps + 1, now, c * (i + 1), i);
    }
}
int main() {
    int t;
    cin >> t;
    while (t--) {
        ans = 0;
        n = read();
        dfs(0, 1, 1, 64);
        cout << ans << endl;
    }
    return 0;
}