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 n;
int LIM = 0;
long long mem[64];
vector<long long> pows[64];
long long rec(int st) {
  if (mem[st] != -1) return mem[st];
  long long ret, sum = 0;
  if (st == 1)
    ret = n;
  else if (st == 2) {
    ret = sqrt(n);
    while (ret * ret > n) ret--;
    while ((ret + 1) * (ret + 1) <= n) ret++;
  } else {
    int ok = 0, not_ok = pows[st].size(), av;
    while (not_ok - ok > 1) {
      av = (ok + not_ok) / 2;
      if (pows[st][av] <= n)
        ok = av;
      else
        not_ok = av;
    }
    ret = ok;
  }
  ret--;
  for (int i = 2; i * st <= LIM; i++) sum = sum + rec(i * st);
  ret = ret - sum;
  mem[st] = ret;
  return ret;
}
int main() {
  for (int st = 3; st < 64; st++) {
    pows[st].push_back(0);
    for (int i = 1; true; i++) {
      long long now = 1e18;
      for (int j = 1; j <= st; j++) now = now / i;
      if (now == 0) break;
      long long prod = 1;
      for (int j = 1; j <= st; j++) prod = prod * i;
      pows[st].push_back(prod);
    }
  }
  int t;
  scanf("%i", &t);
  for (int i = 1; i <= t; i++) {
    for (int j = 0; j < 64; j++) mem[j] = -1;
    scanf("%lld", &n);
    LIM = 0;
    while ((1LL << LIM) <= n) LIM++;
    printf("%lld\n", rec(1));
  }
  return 0;
}
