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<bits/stdc++.h>
#define ll long long
#define endl "\n"
using namespace std;

void solve()
{
	ll n;
	cin>>n;

	// after solving a=sqrt(2c-1)      c can go max n,   2n-1 is odd , ie a is odd ka sqrt

	// 1 is the corner case 

	ll c=0;

	//solution in O(n)
	// for(ll i=3;i*i<=(2*n-1);i+=2)
	// {
	// 	c++;
	// }

	// solution in O(1)

	ll max_odd=sqrt(2*n-1);   //max odd number that a can have
	//corner case is 1;
	// so
	max_odd=max_odd-1;

	cout<<(max_odd/2)<<endl;	//gives number of odds excluding one from max odd.


	// cout<<c<<endl;		
}

int main()
{
  #ifndef ONLINE_JUDGE
  freopen("input.txt","r",stdin);
  freopen("output.txt","w",stdout);
  #endif

	ll test;
	cin>>test;

	while(test--)
	{
		solve();
	}
return 0;
}