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;
long long n, T, f[100], kkk = 1e18, ff[100];
long long pd1(long long x, long long y) {
  long long ans = 1;
  for (long long i = 1; i <= x; i++)
    if (ans > (double)kkk / y)
      return 0;
    else
      ans *= y;
  return 1;
}
void init() {
  for (long long i = 2; i < 100; i++) {
    long long s = 0, k = i;
    for (long long j = 2; j <= k; j++)
      if (k % j == 0) {
        if (k % (j * j) == 0) {
          s = -1;
          break;
        }
        s++;
        k /= j;
      }
    if (s < 0)
      f[i] = 0;
    else if (s & 1)
      f[i] = 1;
    else
      f[i] = -1;
  }
  ff[1] = 1e18;
  ff[2] = 1e9;
  ff[3] = 1e6;
  ff[4] = 31622;
  ff[5] = 3981;
  for (long long i = 6; i <= 60; i++)
    for (long long j = 1; j; j++)
      if (!pd1(i, j)) {
        ff[i] = j - 1;
        break;
      }
}
long long ksm(long long x, long long y) {
  if (!y) return 1;
  long long z = ksm(x, y / 2);
  z *= z;
  if (y & 1) z *= x;
  return z;
}
long long pd(long long x, long long y, long long z) {
  if (y > ff[z]) return 0;
  if (ksm(y, z) <= x) return 1;
  return 0;
}
long long dfs(long long x) {
  long long ans = x - 1;
  for (long long i = 2; i <= 60; i++) {
    long long k = (long long)(pow((double)x, (double)1.0 / i));
    k++;
    while (!pd(x, k, i)) k--;
    if (k < 2) break;
    ans -= f[i] * (k - 1);
  }
  return ans;
}
signed main() {
  init();
  scanf("%lld", &T);
  while (T--) {
    scanf("%lld", &n);
    printf("%lld\n", dfs(n));
  }
}
