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:
# c = a^2 - b          a <= b <= c
# c^2 = a^2 + b^2
# c^2 = a^4 - 2a^2b + b^2
# 1 + 2b = a^2 
# c^2 = b**2 + 2b + 1 = (b + 1)^2 => c = b + 1
# a <= b <= b + 1
# b = (a**2 - 1) / 2
# c = ((a**2 - 1) / 2) + 1 =====> ((a**2 - 1) / 2) + 1 <= n
#   <= 2*(n - 1) + 1
# 60k < a^2
def solve():
    n = int(input())
    l, r = 0, int(1e9)
    while r - l > 1:
        m = (r + l)//2
        if m*m <= (n - 1)*2 + 1:
            l = m
        else:
            r = m
    print((l - 1)//2)
    return 
    ans = 0
    for a in range(3, 64000, 2):
        qa = a*a
        if qa & 1 == 0: continue
        b = (qa - 1) // 2
        c = b + 1
        if c > n: break
        if c*c == b*b + qa:
            ans += 1
    print(ans)    
    
    
#d = {x*x: True for x in range(1, 32000) }
t = int(input())
while t > 0:
    t -= 1
    solve()