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 = 1ll << 62;
struct Query {
  long long n;
  int id;
  Query(long long _n = 0, int _id = 0) : n(_n), id(_id) {}
  inline bool operator<(const Query& rhs) const { return n > rhs.n; }
} qry[100010];
long long ans[100010];
vector<pair<long long, int> > dic[65];
inline long long qpow(long long base, int pw) {
  long long res = 1;
  while (pw) {
    if (pw & 1) {
      if (res < inf / base)
        res = res * base;
      else
        return inf;
    }
    if (pw > 1) {
      if (base < inf / base)
        base = base * base;
      else
        return inf;
    }
    pw >>= 1;
  }
  return res;
}
inline long long get_sqrt(long long x) {
  long long l = 1, r = 1000000001ll;
  while (r - l > 1) {
    long long mid = (l + r) >> 1;
    if (mid * mid <= x)
      l = mid;
    else
      r = mid;
  }
  return l;
}
bool notPrime[65];
int mu[65];
vector<int> prime;
void linear_sieve() {
  notPrime[0] = notPrime[1] = true;
  for (int i = 2; i <= 60; i++) {
    if (!notPrime[i]) {
      prime.push_back(i);
      mu[i] = -1;
    }
    for (int j = 0; j < (int)prime.size() && i * prime[j] <= 60; j++) {
      int d = i * prime[j];
      notPrime[d] = true;
      mu[d] = -mu[i];
      if (i % prime[j] == 0) {
        mu[d] = 0;
        break;
      }
    }
  }
}
int main() {
  int T;
  long long n;
  scanf("%d", &T);
  for (int i = 1; i <= T; i++) {
    scanf("%lld", &n);
    qry[i] = Query(n, i);
  }
  sort(qry + 1, qry + 1 + T);
  for (int i = 3; i <= 60; i++) {
    long long v;
    for (int j = 2; (v = qpow(j, i)) <= 1000000000000000000ll; j++)
      dic[i].push_back(pair<long long, int>(v, j));
  }
  linear_sieve();
  for (int i = 1; i <= T; i++) {
    n = qry[i].n;
    long long& res = ans[qry[i].id];
    res = n - 1 + (get_sqrt(n) - 1) * mu[2];
    for (int i = 3; i <= 60; i++) {
      while (!dic[i].empty() && dic[i].back().first > n) dic[i].pop_back();
      if (dic[i].empty()) break;
      res += (dic[i].back().second - 1) * mu[i];
    }
  }
  for (int i = 1; i <= T; i++) printf("%lld\n", ans[i]);
  return 0;
}
