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 = 1e6 + 7;
const int MX = 1e9 + 7;
const long long int INF = 1e18 + 9LL;
long long int n;
long long int dp[100];
vector<long long int> maxi = {
    1,  INF - 9, MX - 7, N - 7, 31622, 3981, 1000, 372, 177, 100, 63, 43, 31,
    24, 19,      15,     13,    11,    10,   8,    7,   7,   6,   6,  5,  5,
    4,  4,       4,      4,     3,     3,    3,    3,   3,   3,   3,  3,  2,
    2,  2,       2,      2,     2,     2,    2,    2,   2,   2,   2,  2,  2,
    2,  2,       2,      2,     2,     2,    2,    2,   2,   2,   2,  2,  2,
    2,  2,       2,      2,     2,     2,    2,    2,   2,   2,   2,  2,  2,
    2,  2,       2,      2,     2,     2,    2,    2};
long long int fast(long long int a, int b) {
  long long int ret = 1LL;
  while (b) {
    if (b & 1) ret *= a;
    b >>= 1;
    a *= a;
  }
  return ret;
}
long long int get(long long int from, long long int to, int pt) {
  to = min(to, maxi[pt]);
  while (from < to) {
    long long int m = (from + to + 1) / 2LL;
    if (fast(m, pt) <= n)
      from = m;
    else
      to = m - 1;
  }
  return from;
}
void solve() {
  scanf("%lld", &n);
  dp[1] = n - 1;
  for (int i = 2; i <= 60; ++i) dp[i] = get(1, dp[i - 1] + 1, i) - 1;
  for (int i = 60; i >= 1; --i)
    for (int j = i + i; j <= 60; j += i) dp[i] -= dp[j];
  printf("%lld\n", dp[1]);
}
int main() {
  int quest;
  scanf("%d", &quest);
  while (quest--) solve();
  return 0;
}
