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 long long inf = (long long)(1e18) + 500;
const long long INF = (long long)1 << 31;
const long double eps = 1e-8;
using namespace std;
int pri[45005];
bool isp[45005];
int mu[45005];
void init(int n) {
  int p = 0;
  mu[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (isp[i] == 0) pri[++p] = i, mu[i] = -1;
    for (int j = 1; j <= p && i * pri[j] <= n; j++) {
      isp[i * pri[j]] = 1;
      if (i % pri[j] == 0) {
        mu[i * pri[j]] = 0;
        break;
      } else
        mu[i * pri[j]] = -mu[i];
    }
  }
}
long long quick_pow(long long a, long long b) {
  long long res = 1, tmp = a;
  while (b) {
    if (b & 1) {
      long double judge = 1.0 * inf / res;
      if (tmp > judge) return -1;
      res *= tmp;
    }
    b >>= 1;
    if (tmp > INF && b > 0) return -1;
    tmp = tmp * tmp;
  }
  return res;
}
long long find(long long a, long long b) {
  long long r = (long long)(powl((long double)(a), (long double)1.0 / b) + eps);
  long long p = quick_pow(r, b);
  if (p == a) return r;
  if (p == -1 || p > a) return r - 1;
  long long t = quick_pow(r + 1, b);
  if (t != -1 && t <= a)
    return r + 1;
  else
    return r;
}
int main() {
  int T;
  scanf("%d", &T);
  init(1000);
  while (T--) {
    long long n;
    long long ans = 0;
    scanf("%I64d", &n);
    for (int i = 1; i <= 100; i++) {
      if (mu[i] == 0) continue;
      ans += mu[i] * (find(n, i) - 1);
    }
    printf("%lld\n", ans);
  }
}
