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 JAVA):

Solution:
import java.util.*;

public class D {
    static long pythagoreanTriplets(long limit) {

        long a, b, c = 0;
        long m = 2;
        long count = 0;
        while (c < limit) {
            int n = 1;
            while (n < m) {
                a = m * m - (long) n * n;
                b = 2 * m * n;
                c = m * m + (long) n * n;
                if (c > limit)
                    break;
                //System.out.println(a + " " + b + " " + c);
                if (((a * a) - b) == c) {
                    count++;
                }
                n++;
            }
            m++;
        }
        return count;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        long t = sc.nextLong();
        while (t-- > 0) {
            long n = sc.nextLong();
            long x = (long) Math.sqrt((double) n/2);
            while (2*x*(x+1)+1 <= n) {
                x = x + 1;
            }
            System.out.println(x-1);

        }
    }
}