Consider some positive integer x. Its prime factorization will be of form x = 2^{k_1} ⋅ 3^{k_2} ⋅ 5^{k_3} ⋅ ...

Let's call x elegant if the greatest common divisor of the sequence k_1, k_2, ... is equal to 1. For example, numbers 5 = 5^1, 12 = 2^2 ⋅ 3, 72 = 2^3 ⋅ 3^2 are elegant and numbers 8 = 2^3 (GCD = 3), 2500 = 2^2 ⋅ 5^4 (GCD = 2) are not.

Count the number of elegant integers from 2 to n.

Each testcase contains several values of n, for each of them you are required to solve the problem separately.

Input

The first line contains a single integer T (1 ≤ T ≤ 10^5) — the number of values of n in the testcase.

Each of the next T lines contains a single integer n_i (2 ≤ n_i ≤ 10^{18}).

Output

Print T lines — the i-th line should contain the number of elegant numbers from 2 to n_i.

Example

Input

4
4
2
72
10


Output

2
1
61
6

Note

Here is the list of non-elegant numbers up to 10:

  * 4 = 2^2, GCD = 2; 
  * 8 = 2^3, GCD = 3; 
  * 9 = 3^2, GCD = 2. 



The rest have GCD = 1.
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
const int inf = 1000000000;
const int maxn = 1000005;
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int INF = 0x3f3f3f3f;
clock_t start, finish;
double duration;
long long prime[maxn], mob[maxn], vis[maxn], cnt;
int counter[maxn];
void Mobius() {
  mob[1] = 1;
  cnt = 0;
  for (long long i = 2; i < maxn; i++) {
    counter[i] = counter[i - 1];
    if (!vis[i]) {
      prime[cnt++] = i;
      mob[i] = -1;
      counter[i]++;
    }
    for (long long j = 0; j < cnt && i * prime[j] < maxn; j++) {
      vis[i * prime[j]] = 1;
      if (i % prime[j])
        mob[i * prime[j]] = -mob[i];
      else {
        mob[i * prime[j]] = 0;
        break;
      }
    }
  }
}
int rt_p[65];
long long MAXX = 1e18;
long long b[100005], a[100005], rt[100][4200], sq3[1000005];
map<long long, long long> mp;
int main() {
  Mobius();
  long long n;
  int T;
  ios::sync_with_stdio(0);
  cin >> T;
  while (T--) {
    cin >> n;
    long long ans = n - 1;
    for (int i = 2; i <= 60; i++) {
      long long t = (long double)powl(n + 0.1, 1.0 / i) - 1;
      if (t == 0) break;
      ans += t * mob[i];
    }
    cout << ans << endl;
  }
}
