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 ll = long long;
using namespace std;
const ll N = 60, MAX = 1e18;
int t, s[N], log_s[N];
map<ll, bool> m;
vector<ll> ans;
ll binpow(ll x, int k) {
  ll res = 1;
  while (k) {
    if (k & 1) {
      if (x && MAX / x < res) {
        return 0;
      }
      res = res * x;
    }
    if (x && MAX / x < x) {
      x = 0;
    }
    x = x * x;
    k >>= 1;
  }
  return res;
}
ll sqrt2(ll n) {
  ll x = sqrt(n) + 0.5;
  if (x * x > n) x--;
  return x;
}
ll solve(ll L, ll R) {
  return upper_bound(ans.begin(), ans.end(), R) -
         lower_bound(ans.begin(), ans.end(), 2) + (sqrt2(R) - sqrt2(L - 1));
}
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  cin >> t;
  ll L = 2, R = 1e18;
  for (int i = 3; i < 60; i++) {
    int l = 1e6, r = 1;
    for (int j = 20; j >= 0; j--) {
      ll x = binpow(l - (1 << j), i), y = binpow(r + (1 << j), i);
      if (l - (1 << j) > 1 && (!x || x >= L)) {
        l -= 1 << j;
      }
      if (y && r + (1 << j) <= 1e6 && y <= R) {
        r += 1 << j;
      }
    }
    for (int j = l; j <= r; j++) {
      ll x = binpow(j, i);
      if (!m[x] && sqrt2(x) * sqrt2(x) != x) {
        ans.push_back(x);
        m[x] = 1;
      }
    }
  }
  sort(ans.begin(), ans.end());
  while (t--) {
    ll n;
    cin >> n;
    cout << n - solve(2, n) - 1 << '\n';
  }
}
