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 ios ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
#define pb push_back
#define po pop_back
#define YES cout<<"Yes\n"
#define NO cout<<"No\n"
#define For(i,a,b) for (int i = a; i <= b; i++)
#define FFOR(i,a,b) for (int i = a; i >= b; i--)
const int ma = 1e5+10;
using namespace std;
int a[ma];
void solve()
{
	int n; cin>>n;
	int sum = 0;
	for(int i=3;i<n;i+=2)
	{
		if(i*i/2+1<=n)
			sum++;
		else break;
	}
	cout<<sum<<'\n';
	return;
}
signed main()
{
	ios;
	int tt; cin>>tt;
	while(tt--)
	{
		solve();
	}
}