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;

int main() {
    int t;
    cin >> t;
    
    while (t --> 0) {
        int n;
        cin >> n;
        
        /*
        c = a^2 - b 
        c^2 = a^2 + b^2
        
        c + b = (c^2 - b^2)
        c + b = (c - b) * (c + b)
        c - b = 1
        c = b + 1
        
        b + 1 = a^2 - b
        2 * b + 1 = a^2 (a is odd)
        
        b = (a^2 - 1) / 2
        c = (a^2 + 1) / 2
        
        
        a > 1: (1^2 - 1) / 2 = 0 - fail
        
        a <= b <= c <= n
        
        (a^2 + 1) / 2 <= n
        
        a^2 <= 2 * n - 1 (!)
        */
        
        int lim = 2 * n - 1;
        
        int a = sqrt(lim);
        while (a * a > lim) --a;
        while ((a + 1) * (a + 1) <= lim) ++a;
        
        int ans = (a - 1) / 2;
        
        cout << ans << endl;
    }
}