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;
const int N = 100002;
int t;
long long n, dp[70];
long long power(long long x, int p) {
  long long ret = 1;
  while (not false) {
    if (p & 1) {
      if (ret > n / x) return -1;
      ret *= x;
    }
    p /= 2;
    if (!p) return ret;
    if (x > n / x) return -1;
    x *= x;
  }
}
int main() {
  dp[1] = 1000000000;
  scanf("%d", &t);
  while (t--) {
    scanf("%I64d", &n);
    for (int i = 2; i <= 63; ++i) {
      int l = 1, r = dp[i - 1] + 1;
      while (l < r) {
        int mid = (l + r) / 2 + 1;
        if (~power(mid, i))
          l = mid;
        else
          r = mid - 1;
      }
      dp[i] = l - 1;
      if (!dp[i])
        while (i < 63) dp[++i] = 0;
    }
    long long ans = 0;
    for (int i = 63; i >= 2; --i) {
      for (int j = 2 * i; j <= 63; j += i) dp[i] -= dp[j];
      ans += dp[i];
    }
    printf("%I64d\n", n - 1 - ans);
  }
}
