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;
int ar[61] = {}, tp, ps, j, cr;
long long a;
long long pw(long long b, long long e) {
  long long t = 1;
  while (e) {
    if (e & 1) {
      t *= b;
    }
    b *= b;
    e >>= 1;
  }
  return t;
}
double estmpw(double b, long long e) {
  double t = 1;
  while (e) {
    if (e & 1) {
      t *= b;
    }
    b *= b;
    e >>= 1;
  }
  return t;
}
long long hmp(long long x, long long p) {
  long long l = 2, r = pow(x, (double)1 / p) + 5;
  while (l < r) {
    long long pv = (l + r) >> 1;
    if (estmpw(pv, p) > 5e18) {
      r = pv;
    } else if (pw(pv, p) <= x) {
      l = pv + 1;
    } else {
      r = pv;
    }
  }
  return max((long long)0, min(l, r) - 2);
}
long long clc(long long x) {
  long long to = 0;
  for (long long i = 2; i <= 60; i++) {
    if (ar[i] == 1) {
      to += hmp(x, i);
    } else if (ar[i] == 2) {
      to -= hmp(x, i);
    }
  }
  return x - to - 1;
}
int main() {
  for (int i = 2; i <= 60; i++) {
    tp = i, ps = 0;
    for (j = 2; j * j <= tp; j++) {
      cr = 0;
      while (tp % j == 0) {
        cr++;
        tp /= j;
      }
      if (cr > 1) {
        ps = -1;
        break;
      } else if (cr == 1) {
        ps++;
      }
    }
    if (ps != -1) {
      if (tp > 1) {
        ps++;
      }
      if (ps & 1) {
        ar[i] = 1;
      } else {
        ar[i] = 2;
      }
    }
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    long long a;
    scanf("%lld", &a);
    printf("%lld\n", clc(a));
  }
  return 0;
}
