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 PYTHON3):

Solution:
import io
import os

from collections import Counter, defaultdict, deque

from bisect import bisect_left, bisect_right

oddSquares = []  # exclude 1
for i in range(3, int(10 ** 5) + 1, 2):
    oddSquares.append(i * i)

assert 2 * (10 ** 9) - 1 <= oddSquares[-1]


def solveBrute(N,):
    count = 0
    for c in range(1, N + 1):
        aa = 2 * c - 1
        if aa in oddSquares:
            count += 1

            b = c - 1
            a = int(aa ** 0.5)
            assert 1 <= a <= b <= c <= N
            assert a * a + b * b == c * c
            assert c == a * a - b

    return count


def solve(N,):
    count = bisect_right(oddSquares, 2 * N - 1)
    return count


if False:
    for N in range(1, 100):
        ans, ans1 = solve(N), solveBrute(N)
        print(N, ans, ans1)
        assert ans == ans1

if __name__ == "__main__":
    input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline

    TC = int(input())
    for tc in range(1, TC + 1):
        (N,) = [int(x) for x in input().split()]
        ans = solve(N,)
        print(ans)