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 maxn = 65;
const long long INF64 = 3e18;
int primes[maxn], top, is[maxn], mu[maxn];
long long ans[100005];
int cnt[61];
void seive() {
  fill(mu, mu + maxn, 1);
  mu[0] = 0;
  for (int i = 2; i < maxn; i++) {
    if (!is[i]) {
      primes[top++] = i;
      mu[i] = -1;
    }
    for (int j = 0; j < top && primes[j] * i < maxn; j++) {
      is[primes[j] * i] = 1;
      if (i % primes[j] == 0) {
        mu[primes[j] * i] = 0;
        break;
      } else {
        mu[primes[j] * i] = -mu[i];
      }
    }
  }
}
long long mypow(long long a, int b) {
  long long ret = 1;
  long long c = a;
  while (b) {
    if (b & 1) {
      if (ret < INF64 / c)
        ret *= c;
      else
        return INF64;
    }
    b >>= 1;
    if (b) {
      if (c < INF64 / c)
        c *= c;
      else
        return INF64;
    }
  }
  return ret;
}
int work1(long long n, int h) {
  int left = 1;
  int right = 1e6;
  if (h == 2) right = 1e9;
  while (left < right) {
    int mid = (left + right + 1) / 2;
    if (mypow(mid, h) <= n)
      left = mid;
    else
      right = mid - 1;
  }
  return left;
}
int main() {
  seive();
  int T;
  scanf("%d", &T);
  vector<pair<long long, int>> v;
  for (int i = 1; i <= T; i++) {
    long long temp;
    cin >> temp;
    ans[i] = temp - 1 - work1(temp, 2) - work1(temp, 3) - work1(temp, 5) + 3;
    v.push_back({temp, i});
  }
  sort(v.begin(), v.end(), greater<pair<long long, int>>());
  for (int i = 6; i <= 60; i++) cnt[i] = 1000;
  int lim = 60;
  for (auto p : v) {
    for (int i = 6; i <= lim; i++) {
      while (mypow(cnt[i], i) > p.first) cnt[i]--;
      if (cnt[i] == 1) lim = i - 1;
    }
    for (int i = 6; i <= lim; i++) ans[p.second] += mu[i] * (cnt[i] - 1);
  }
  for (int i = 1; i <= T; i++) cout << ans[i] << endl;
  return 0;
}
