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;
bool flag[66];
int d[66];
int MAX[66];
bool prime(int x) {
  for (int i = 2; i <= x - 1; i++)
    if (x % i == 0) return false;
  return true;
}
long long pow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans *= a;
    a *= a;
    b >>= 1;
  }
  return ans;
}
long long calc(int x, long long y) {
  long long ans = 0;
  long long l = 1;
  long long r = MAX[x];
  while (l + 1 < r) {
    long long mid = (l + r) >> 1;
    long long res = y;
    res /= pow(mid, x);
    if (res)
      l = mid;
    else
      r = mid;
  }
  ans = l;
  return ans;
}
long long solve(long long n) {
  long long ans = 0;
  for (int i = 2; i <= 64; i++) {
    if (d[i]) ans += d[i] * calc(i, n);
  }
  return ans;
}
int main() {
  for (int i = 2; i <= 64; i++) flag[i] = prime(i);
  for (int i = 2; i <= 64; i++)
    if (flag[i]) d[i] = 1;
  for (int i = 2; i <= 64; i++)
    for (int j = i + 1; j <= 64; j++)
      if (flag[i] && flag[j] && i * j <= 64) d[i * j] = -1;
  for (int i = 2; i <= 64; i++)
    for (int j = i + 1; j <= 64; j++)
      for (int k = j + 1; k <= 64; k++)
        if (flag[i] && flag[j] && flag[k] && i * j * k <= 64) d[i * j * k] = 1;
  for (int i = 2; i <= 64; i++) {
    if (d[i]) {
      long long l = 1;
      long long r = 1e18;
      long long y = 1e18;
      while (l + 1 < r) {
        long long mid = (l + r) >> 1;
        long long res = y;
        for (int j = 1; j <= i; j++) res /= mid;
        if (res)
          l = mid;
        else
          r = mid;
      }
      MAX[i] = r;
    }
  }
  int T;
  cin >> T;
  while (T--) {
    long long n;
    scanf("%lld", &n);
    long long ans = n + 1 - solve(n);
    printf("%lld\n", ans);
  }
  return 0;
}
