#include <bits/stdc++.h>
using namespace std;
typedef unsigned long long ull;
#define rep(i, j, k) for (int i = (j); i <= (k); ++i)
#define per(i, j, k) for (int i = (j); i >= (k); --i)
#define sz(v) int((v).size())
#define all(v) (v).begin(), (v).end()
#define endl "\n"
#define no cout << "NO" << endl;
#define yes cout << "YES" << endl;
using ll = long long;
using pii = std::pair<int, int>;
using pll = std::pair<ll, ll>;
typedef double db;
const db eps = 1e-8;
#define inf 1e18
#define ret                 \
    {                       \
        cout << -1 << endl; \
        return;             \
    }
template <typename T>
bool gmax(T &x, const T y)
{
    if (x < y)
        return x = y, 1;
    return 0;
}
template <typename T>
bool gmin(T &x, const T y)
{
    if (y < x)
        return x = y, 1;
    return 0;
}
#define maxn 200003
ll isn_prime[maxn], primes[1000000], pnum = 0;
void init()
{
    isn_prime[0] = isn_prime[1] = 1;
    rep(i, 2, maxn - 3)
    {
        if (!isn_prime[i])
            primes[++pnum] = i;
        for (int j = 1; j <= pnum && primes[j] * i <= maxn - 3; ++j)
        {
            isn_prime[primes[j] * i] = 1;
            if (i % primes[j] == 0)
                break;
        }
    }
}
void solve()
{
    ll x, cnt1 = 0, cnt2 = 0, j = 1;
    cin >> x;
    if (x == 1)
    {
        cout << 1 << endl;
        return;
    }
    while (x != 1 && j < 1e5 + 1 && primes[j] <= x)
    {
        int k = 0;
        while (x % primes[j] == 0)
        {
            k++;
            cnt1++;
            x /= primes[j];
        }
        if (k)
            cnt2++;
        j++;
    }
    if (x != 1 || cnt2 == 1)
    {
        cout << cnt1 + 1 << endl;
        return;
    }
    cout << (1ll * cnt2 * cnt2 - cnt2) / 2 + 2 + cnt1 << endl;
}
signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t = 1;
    init();
    cin >> t;
    while (t--)
        solve();
}
// 循环边界
// 二分l和r的范围
// 数组越界
// 在函数内开数组赋初值
// long long
// 尽量用数学语言刻画
// 从直觉到严谨分析，推导
// 冷静，冷静，冷静
//  if(mp[x]) 会生成新的键值