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 <stdio.h>
#include <math.h>
#include<bits/stdc++.h>
using namespace std;


// C = a^2 - b and C^2 = a^2 + b^2 hence (a^2 - b)^2 = a^2 + b^2 , evaluate and subtract then, a^2 = 2*b+1
// Since a^2 - b = C , then 2*b+1 - b = C , then C = b+1 !!!!! , We can observe that the difference between b and c are just 1
// b and c are nearly equal , so we must find integer "a" so that its square is odd number since odd / 2 gives you two numbers
// the floor and the ceil , which are basically b and b+1 -> b and c ^^

int main() {


    int t;

    scanf("%d",&t);

    while(t--) {

        int n;

        scanf("%d",&n);

        int odd_Square = sqrt(n << 1); // --> sqrt ( n / 2 );

        printf("%d\n",(odd_Square-1) >> 1); // --> (odd_Square - 1) / 2 ;

    }

    return 0;

}