#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
long long a[N], b[1050], c[N];
int cnt = 0;
void init() {
  cnt = 0;
  for (int i = 1; i < N - 50; i++) a[cnt++] = i * i;
  cnt = 0;
  for (int i = 1; i < 1010; i++) b[cnt++] = i * i * i;
  int u = 0, t = 0;
  cnt = 0;
  while (a[u] && b[t]) {
    if (a[u] < b[t]) {
      c[cnt++] = a[u++];
    } else if (a[u] == b[t]) {
      c[cnt++] = a[u++];
      t++;
    } else
      c[cnt++] = b[t++];
  }
}
int main() {
  init();
  int T;
  cin >> T;
  while (T--) {
    int n;
    scanf("%d", &n);
    int i = 0;
    while (c[i] <= n) {
      i++;
    }
    printf("%d\n", i);
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
const int N = 1e5 + 10;
long long a[N], b[1050], c[N];
int cnt = 0;
void init() {
  cnt = 0;
  for (int i = 1; i < N - 50; i++) a[cnt++] = i * i;
  cnt = 0;
  for (int i = 1; i < 1010; i++) b[cnt++] = i * i * i;
  int u = 0, t = 0;
  cnt = 0;
  while (a[u] && b[t]) {
    if (a[u] < b[t]) {
      c[cnt++] = a[u++];
    } else if (a[u] == b[t]) {
      c[cnt++] = a[u++];
      t++;
    } else
      c[cnt++] = b[t++];
  }
}
int main() {
  init();
  int T;
  cin >> T;
  while (T--) {
    int n;
    scanf("%d", &n);
    int i = 0;
    while (c[i] <= n) {
      i++;
    }
    printf("%d\n", i);
  }
  return 0;
}

```