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;
inline long long intpow(long long x, int p, long long bnd) {
  long long res = 1;
  while (p) {
    if (p & 1) {
      if (res > bnd / x) return -1;
      res *= x;
    }
    if (p > 1) {
      if (x > bnd / x) return -1;
      x *= x;
    }
    p >>= 1;
  }
  return res;
}
inline long long binsearch_root(int pw, long long bnd) {
  long long L = 1, R = (long long)1e9 + 2;
  while (R - L > 1) {
    long long mid = (L + R) / 2;
    if (intpow(mid, pw, bnd) == -1)
      R = mid;
    else
      L = mid;
  }
  return L;
}
int mu[100];
long long maxpow[64];
long long solve(long long n) {
  unsigned long long ans = 0;
  for (int pw = 2; pw <= 59; pw++) {
    long long maxnum;
    if (pw > 2) {
      while (intpow(maxpow[pw], pw, n) == -1) --maxpow[pw];
      maxnum = maxpow[pw];
    } else {
      maxnum = binsearch_root(2, n);
    }
    ans += mu[pw] * (maxnum - 1);
  }
  return (long long)(n - 1 - ans);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  for (int i = 2; i <= 59; i++) {
    int j = i;
    mu[i] = -1;
    for (int d = 2; d <= j; d++) {
      if (j % d == 0) {
        mu[i] *= -1;
        j /= d;
        if (j % d == 0) {
          mu[i] = 0;
          break;
        }
      }
    }
  }
  for (int pw = 3; pw <= 59; pw++)
    maxpow[pw] = binsearch_root(pw, (long long)1e18);
  int T;
  scanf("%d", &T);
  vector<pair<long long, int> > avtor_zadachi_pidoras;
  for (int i = 0; i < T; i++) {
    long long n;
    scanf("%lld", &n);
    avtor_zadachi_pidoras.push_back(make_pair(n, i));
  }
  sort((avtor_zadachi_pidoras).begin(), (avtor_zadachi_pidoras).end());
  reverse((avtor_zadachi_pidoras).begin(), (avtor_zadachi_pidoras).end());
  vector<long long> ans(T);
  for (auto query : avtor_zadachi_pidoras) {
    ans[query.second] = solve(query.first);
  }
  for (int i = 0; i < T; i++) printf("%lld\n", ans[i]);
  return 0;
}
