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 N = 64;
bool isPrime[N];
int mu[N], T;
long long n;
vector<int> prime;
const long long INF = 1e18 + 300;
const long long TT = (long long)1 << 31;
void MobiwusiInit() {
  memset(isPrime, true, sizeof(isPrime));
  isPrime[1] = false;
  mu[1] = 1;
  for (int i = 2; i < N; i++) {
    if (isPrime[i]) {
      prime.push_back(i);
      mu[i] = -1;
    }
    for (int j = 0; j < prime.size() && i * prime[j] < N; j++) {
      isPrime[i * prime[j]] = false;
      if (i % prime[j] == 0) {
        mu[i * prime[j]] = 0;
        break;
      }
      mu[i * prime[j]] = -mu[i];
    }
  }
}
long long multi(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) {
      double judge = 1.0 * INF / ans;
      if (a > judge) return -1;
      ans *= a;
    }
    b >>= 1;
    if (a > TT && b > 0) return -1;
    a = a * a;
  }
  return ans;
}
long long find(long long x, long long k) {
  long long r = (long long)powl(x, 1.0 / k);
  long long t, p;
  p = multi(r, k);
  if (p == x)
    return r;
  else if (p > x || p == -1) {
    do {
      r--;
      p = multi(r, k);
    } while (p > x || p == -1);
  } else {
    t = multi(r + 1, k);
    while (t != -1 && t <= x) {
      r++;
      t = multi(r + 1, k);
    }
  }
  return r;
}
int main() {
  MobiwusiInit();
  scanf("%d", &T);
  while (T--) {
    long long sum = 0;
    scanf("%lld", &n);
    for (int i = 1; i < N; i++) {
      if (!mu[i]) continue;
      long long tmp = find(n, i);
      if (tmp <= 1) break;
      sum += mu[i] * (tmp - 1);
    }
    printf("%lld\n", sum);
  }
  return 0;
}
