#include <bits/stdc++.h>
using namespace std;
typedef long long int ll; 

#define debug(x) cout << #x << " " << x << endl

#define debug1(x, y, z) \
  cout << #x << " " << x << endl << #y << " " << y << endl << #z << " " << z << endl

#define debugarr(arr, N)      \
  cout << #arr << " ";        \
  for (int i = 0; i < N; i++) \
    cout << arr[i] << " ";    \
  cout << endl;

#define pi 3.1415926535897932384626433832

#define endl "\n"

bool sortcol(const vector<ll> &v1, const vector<ll> &v2)
{
  return v1[1] < v2[1];
}

void solve()
{
  ll N;
  cin >> N;
  ll i = sqrt(2*N-1);
  cout<<(i-1)/2<<endl;
}
int main()
{
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ll t = 1;
  cin >> t;
  while (t--)
  {
    solve();
  }
}

/*
_________________ IMP NOTES __________________
a = 97
A = 65


_________________To Do______________________
add debug stmts
add fast io
print whole array
*/
### Prompt

Please create a solution in CPP to the following problem:

A Pythagorean triple is a triple of integer numbers (a, b, c) such that it is possible to form a right triangle with the lengths of the first cathetus, the second cathetus and the hypotenuse equal to a, b and c, respectively. An example of the Pythagorean triple is (3, 4, 5).

Vasya studies the properties of right triangles, and he uses a formula that determines if some triple of integers is Pythagorean. Unfortunately, he has forgotten the exact formula; he remembers only that the formula was some equation with squares. So, he came up with the following formula: c = a^2 - b.

Obviously, this is not the right formula to check if a triple of numbers is Pythagorean. But, to Vasya's surprise, it actually worked on the triple (3, 4, 5): 5 = 3^2 - 4, so, according to Vasya's formula, it is a Pythagorean triple.

When Vasya found the right formula (and understood that his formula is wrong), he wondered: how many are there triples of integers (a, b, c) with 1 ≤ a ≤ b ≤ c ≤ n such that they are Pythagorean both according to his formula and the real definition? He asked you to count these triples.

Input

The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.

Each test case consists of one line containing one integer n (1 ≤ n ≤ 10^9).

Output

For each test case, print one integer — the number of triples of integers (a, b, c) with 1 ≤ a ≤ b ≤ c ≤ n such that they are Pythagorean according both to the real definition and to the formula Vasya came up with.

Example

Input


3
3
6
9


Output


0
1
1

Note

The only Pythagorean triple satisfying c = a^2 - b with 1 ≤ a ≤ b ≤ c ≤ 9 is (3, 4, 5); that's why the answer for n = 3 is 0, and the answer for n = 6 (and for n = 9) is 1.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
typedef long long int ll; 

#define debug(x) cout << #x << " " << x << endl

#define debug1(x, y, z) \
  cout << #x << " " << x << endl << #y << " " << y << endl << #z << " " << z << endl

#define debugarr(arr, N)      \
  cout << #arr << " ";        \
  for (int i = 0; i < N; i++) \
    cout << arr[i] << " ";    \
  cout << endl;

#define pi 3.1415926535897932384626433832

#define endl "\n"

bool sortcol(const vector<ll> &v1, const vector<ll> &v2)
{
  return v1[1] < v2[1];
}

void solve()
{
  ll N;
  cin >> N;
  ll i = sqrt(2*N-1);
  cout<<(i-1)/2<<endl;
}
int main()
{
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ll t = 1;
  cin >> t;
  while (t--)
  {
    solve();
  }
}

/*
_________________ IMP NOTES __________________
a = 97
A = 65


_________________To Do______________________
add debug stmts
add fast io
print whole array
*/
```