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 K = 100;
const int N = 100 * 1000 + 13;
const long long INF64 = 3e18;
int mu[K];
void precalc() {
  static bool prime[K];
  static int lst[K];
  memset(prime, false, sizeof(prime));
  for (int i = 0; i < int(K); i++) lst[i] = i;
  for (int i = 2; i < K; ++i) {
    if (lst[i] == i) mu[i] = 1;
    for (int j = 2 * i; j < K; j += i) {
      lst[j] = min(lst[j], lst[i]);
      if (lst[j] == lst[i])
        mu[j] = 0;
      else
        mu[j] = -mu[i];
    }
  }
}
int mx[K];
long long binpow(long long a, int b) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      if (res < INF64 / a)
        res *= a;
      else
        return INF64;
    }
    if (b > 1) {
      if (a < INF64 / a)
        a *= a;
      else
        return INF64;
    }
    b >>= 1;
  }
  return res;
}
long long calc(long long n) {
  int pw = 63 - __builtin_clzll(n);
  for (int i = 3; i <= pw; ++i) {
    if (mu[i] == 0) continue;
    while (binpow(mx[i], i) > n) --mx[i];
  }
  long long res = n - 1;
  for (int i = 2; i <= pw; ++i) res -= mu[i] * (mx[i] - 1);
  return res;
}
int get_sqrt(long long n) {
  int l = 1, r = 1000000000;
  while (l < r - 1) {
    int m = (l + r) / 2;
    if (m * 1ll * m <= n)
      l = m;
    else
      r = m;
  }
  return (r * 1ll * r <= n ? r : l);
}
long long ans[N];
int main() {
  precalc();
  int T;
  scanf("%d", &T);
  vector<pair<long long, int>> q;
  for (int i = 0; i < int(T); i++) {
    long long n;
    scanf("%lld", &n);
    q.push_back({n, i});
  }
  sort(q.begin(), q.end(), greater<pair<long long, int>>());
  mx[3] = 1000000;
  mx[4] = 31622;
  mx[5] = 3981;
  for (int i = 6; i < K; ++i) mx[i] = 1000;
  for (int z = 0; z < int(T); z++) {
    long long n = q[z].first;
    mx[2] = get_sqrt(n);
    ans[q[z].second] = calc(n);
  }
  for (int i = 0; i < int(T); i++) printf("%lld\n", ans[i]);
  return 0;
}
