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 long long inf = 2e18 + 5;
long long fastPow(long long x, int p) {
  long long answer = 1, current = x;
  while (p != 0) {
    if (p % 2 != 0) {
      if (answer > inf / current + 1) return inf;
      answer *= current;
    }
    p /= 2;
    if (current <= inf / current + 1)
      current *= current;
    else
      current = inf;
  }
  return answer;
}
long long dp[70];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int T;
  cin >> T;
  while (T--) {
    long long n;
    cin >> n;
    memset(dp, 0, sizeof(dp));
    dp[1] = n - 1;
    long long lastAns = n;
    for (int p = 2; p < 65; p++) {
      long long l = 1, r = lastAns, mid;
      while (l + 1 < r) {
        mid = (l + r) / 2;
        if (fastPow(mid, p) <= n)
          l = mid;
        else
          r = mid - 1;
      }
      if (fastPow(r, p) <= n)
        lastAns = r;
      else
        lastAns = l;
      dp[p] = lastAns - 1;
    }
    for (int i = 64; i >= 2; i--) {
      for (int d = 1; d * d <= i; d++) {
        if (i % d == 0) {
          dp[d] -= dp[i];
          if (d != 1 && i / d != d) dp[i / d] -= dp[i];
        }
      }
    }
    cout << dp[1] << '\n';
  }
}
