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>
const int D = 61, C = 100001, CC = 1000001;
int f[D], end[D];
long long dp[D][C], d3[CC];
inline long long poww(long long a, long long b) {
  if (b == 0) return 1;
  long long c = 1;
  while (b > 0) {
    if (b % 2 == 1) c = c * a;
    b /= 2;
    a = a * a;
  }
  return c;
}
inline int binlow(long long tab[], long long v, int l, int r) {
  int m;
  while (l <= r) {
    m = (l + r) / 2;
    if (tab[m] > v)
      r = m - 1;
    else
      l = m + 1;
  }
  return r;
}
inline long long bin2(long long v, long long l, long long r) {
  long long m;
  while (l <= r) {
    m = (l + r) / 2;
    if (m * m > v)
      r = m - 1;
    else
      l = m + 1;
  }
  return r;
}
int main() {
  int i, j, q;
  long long a, b, wynn = 0, z;
  double x, y, c;
  for (i = 2; i < D; i++) {
    if (f[i] == 0) {
      f[i] = 1;
      for (j = i + i; j < D; j += i)
        if (f[j] != -2) f[j]++;
      for (j = i * i; j < D; j += (i * i)) f[j] = -2;
    }
  }
  for (i = 2; i < D; i++) {
    if (f[i] == -2)
      f[i] = 0;
    else if (f[i] % 2 == 0)
      f[i] = -1;
    else
      f[i] = 1;
  }
  for (j = 0; j < CC; j++) {
    z = poww(j, 3);
    d3[j] = z;
  }
  end[3] = j - 1;
  for (i = 5; i < D; i++) {
    if (f[i] == 0) continue;
    for (j = 0; j < C; j++) {
      z = poww(j, i);
      if (log10(j) * i > 18.6) break;
      dp[i][j] = z;
    }
    end[i] = j - 1;
  }
  scanf("%d", &q);
  while (q--) {
    scanf("%lld", &a);
    wynn = bin2(a, 0, 1000000000) - 1;
    b = binlow(d3, a, 0, end[3]);
    wynn = wynn + (b - 1);
    for (i = 5; i < 61; i++) {
      if (f[i] == 0) continue;
      b = binlow(dp[i], a, 0, end[i]);
      wynn = wynn + (b - 1) * f[i];
    }
    printf("%lld\n", a - wynn - 1);
  }
  return 0;
}
