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>
const int N = 3e5 + 10;
const int mod = 1e9 + 7;
using namespace std;
int p[66], pr[66], u[66], cnt;
bool isp[66];
void init(int n) {
  u[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (isp[i] == 0) pr[++cnt] = i, u[i] = -1;
    for (int j = 1; j <= cnt && i * pr[j] <= n; j++) {
      int x = i * pr[j];
      isp[x] = 1;
      if (i % pr[j] == 0) {
        u[x] = 0;
        break;
      } else
        u[x] = -u[i];
    }
  }
}
vector<long long> v[62];
long long sq(long long n) {
  long long x = sqrt(n);
  while (x * x < n) x++;
  while (x * x > n) x--;
  return x;
}
long long ok(long long n, long long x) {
  if ((1LL << n) > x) return 1;
  if (n == 2) return sq(x);
  long long q = upper_bound(v[n].begin(), v[n].end(), x) - v[n].begin() + 1;
  return q;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  init(65);
  for (int i = 2; i <= 1000000; i++) {
    long long a = i;
    for (int j = 2; j <= 60; j++) {
      if (1e18 / a < i) break;
      a *= i;
      v[j].push_back(a);
    }
  }
  int t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long ans = n;
    for (int g = 2; g <= 60; g++) {
      long long q = ok(g, n);
      if (q < 2) break;
      ans += u[g] * (q - 1);
    }
    cout << ans - 1 << endl;
  }
  return 0;
}
