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:
#THIS CODE IS MADE BY "harshest2020"
from sys import *
from math import floor,ceil,sqrt
ws=lambda:map(int,stdin.readline().strip().split())
li=lambda:list(map(int,stdin.readline().strip().split()))
mod=1000000007
def ncr(n, r, p): 
    num = den = 1 
    for i in range(r): 
        num = (num * (n - i)) % p 
        den = (den * (i + 1)) % p 
    return (num * pow(den, p - 2, p)) % p
 
def gcd(a,b):
    if (b == 0): 
         return a 
    return gcd(b, a%b)
def prod(l):
    ans=1
    for i in range(len(l)):
        ans=ans*l[i]
    return ans
def sortindex(l,a):
    c=[]
    if(a==-1):
        rev=True
    else:
        rev=False
    for i in range(len(l)):
        c.append([l[i],i])
    x=sorted(c,reverse=rev)
    print(x)
    c=[]
    for i in range(len(l)):
        c.append(x[i][1])
    return c
def calc(n):
    c=sqrt(1+4*n)
    if(c==int(c)):
        return c
    else:
        return -1

for _ in range(int(input())):
    n=int(input())
    if(n<5):
        print(0)
    else:
        ans=0
        for i in range(3,n):
            if(i&1):
                if((i*i+1)/2>n):
                    break
                else:
                    ans+=1
        print(ans)