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;
long long mu[1010], prime[1010];
bool vis[1010];
void getmu() {
  long long cnt = 0;
  mu[1] = 1;
  memset(vis, false, sizeof(vis));
  for (long long i = 2; i < 1010; ++i) {
    if (!vis[i]) {
      prime[cnt++] = i;
      mu[i] = -1;
    }
    for (long long j = 0; j < cnt && i * prime[j] < 1010; ++j) {
      vis[i * prime[j]] = true;
      if (i % prime[j])
        mu[i * prime[j]] = -mu[i];
      else {
        mu[i * prime[j]] = 0;
        break;
      }
    }
  }
}
long long power(long long x, long long y) {
  long long ret = 1;
  for (; y; y >>= 1, x = x * x)
    if (y & 1) ret = ret * x;
  return ret;
}
int main() {
  int t;
  scanf("%d", &t);
  getmu();
  long long n;
  while (t--) {
    long long ans = 0;
    scanf("%I64d", &n);
    long long r = 1e9;
    for (long long i = 2, j = 4; j <= n; i++, j = j * 2) {
      long long l = 1, mid, res;
      while (l <= r) {
        mid = (l + r) >> 1;
        if (power(mid, i - 1) <= n / mid)
          res = mid, l = mid + 1;
        else
          r = mid - 1;
      }
      ans += mu[i] * (res - 1);
    }
    printf("%I64d\n", n - 1 + ans);
  }
}
