#include <bits/stdc++.h>
using namespace std;
int T;
long long n;
long long cnt2;
long long cnt3 = 0;
long long cnt6 = 0;
long long qpow(long long a, long long b) {
  if (b == 0) return 1;
  long long ans = qpow(a, b / 2);
  if ((b & 1))
    return ans * ans * a;
  else
    return ans * ans;
}
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%lld", &n);
    cnt2 = cnt3 = cnt6 = 0;
    cnt2 = (long long)sqrt(n);
    while (qpow(cnt3, 3) <= n) {
      cnt3++;
    }
    cnt3--;
    while (qpow(cnt6, 6) <= n) {
      cnt6++;
    }
    cnt6--;
    printf("%lld\n", cnt2 + cnt3 - cnt6);
  }
  return 0;
}

### Prompt

Please create a solution in CPP to the following problem:

Polycarp likes squares and cubes of positive integers. Here is the beginning of the sequence of numbers he likes: 1, 4, 8, 9, ....

For a given number n, count the number of integers from 1 to n that Polycarp likes. In other words, find the number of such x that x is a square of a positive integer number or a cube of a positive integer number (or both a square and a cube simultaneously).

Input

The first line contains an integer t (1 ≤ t ≤ 20) — the number of test cases.

Then t lines contain the test cases, one per line. Each of the lines contains one integer n (1 ≤ n ≤ 10^9).

Output

For each test case, print the answer you are looking for — the number of integers from 1 to n that Polycarp likes.

Example

Input


6
10
1
25
1000000000
999999999
500000000


Output


4
1
6
32591
32590
23125

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int T;
long long n;
long long cnt2;
long long cnt3 = 0;
long long cnt6 = 0;
long long qpow(long long a, long long b) {
  if (b == 0) return 1;
  long long ans = qpow(a, b / 2);
  if ((b & 1))
    return ans * ans * a;
  else
    return ans * ans;
}
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%lld", &n);
    cnt2 = cnt3 = cnt6 = 0;
    cnt2 = (long long)sqrt(n);
    while (qpow(cnt3, 3) <= n) {
      cnt3++;
    }
    cnt3--;
    while (qpow(cnt6, 6) <= n) {
      cnt6++;
    }
    cnt6--;
    printf("%lld\n", cnt2 + cnt3 - cnt6);
  }
  return 0;
}

```