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 <iostream>
#include <vector>
#include <queue>
#include <algorithm>
#include <string>
#include <set>
#include <chrono>
#include <random>
#include <map>
#include <bitset>
typedef long long ll;
using namespace std;

int bruteforce(int n)
{
    int ans = 0;
    for (int a = 1; a <= n; a++)
    {
        for (int b = a; b <= n; b++)
        {
            for (int c = b; c <= n; c++)
            {
                if (c * c == a * a + b * b && c == a * a - b) ans++;
            }
        }
    }
    return ans;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t;
    cin >> t;
    while (t--)
    {
        ll n;
        cin >> n;
        ll h = 2 * n - 1;
        ll a = sqrt(h);
        while (a * a > h) a--;
        while ((a + 1) * (a + 1) <= h) a++;
        a = min(a, n);
        cout << (a + 1) / 2 - 1 << "\n";
        //cout << bruteforce(n) << "\n";
        //if ((a + 1) / 2 - 1 != bruteforce(n))
    }
    return 0;
}