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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

#define name ""
#define ini freopen(name".inp","r",stdin); freopen(name".out","w",stdout)
#define long long long
#define db double
#define pii pair <int, int>
#define pll pair <long, long>
#define all(c) c.begin(), c.end()
#define fastio ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0)
template <class T> inline void chmin(T &a, const T& val) { if (a > val) a = val; }
template <class T> inline void chmax(T &a, const T& val) { if (a < val) a = val; }
template <class T> long sqr(T a) { return 1ll * a * a; }
template <class T> void compress(T &a){
  sort(all(a)); a.resize(unique(all(a)) - a.begin());  
}

const db pi = acos(-1.00);
const db eps = 1e-6;
const int INF = 1e9;
const int MOD = 1e9+7;
const int dx[4] = {0, 0, -1, 1};
const int dy[4] = {-1, 1, 0, 0};

int main()
{
  fastio;
  int T; cin >> T;
  while (T--){
    int n; cin >> n;
    long ans = 0;
    for (int a = 3; a * a <= 2 * n; a += 2){
      if (a * a / 2 + 1 > n) continue;
      else ans++;
    }
    cout << ans << '\n';
  }
  return 0;
}