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 <iostream>
#include <bits/stdc++.h>
#define fast ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL)
#define ll long long
#define li longll
#define pp pair<ll, ll> 
#define lb lower_bound
#define ub upper_bound
#define mk make_pair
#define pb push_back
#define llp(i, x, n) for (ll i = x; i < n; i++)
#define fr first
#define ss second
#define yes cout << "YES" << endl
#define no cout << "NO" << endl
#define mod 1000000007
using namespace std;
int main()
{
	fast;
	int t;
	cin >> t;
	while (t--)
	{
	   ll n;
	   cin>>n;
	   int c=0;
	   for(ll i=3;i<=100000;i++)
	   {
	   	  	 ll x=i*i+1;
	   	  	 ll y=i*i-1;
	   	  	 if(x%2==0 && y%2==0)
	   	  	 {
	   	  	 	if(x/2<=n && y/2<=n)
	   	  	 	c++;
	   	  	 	else
	   	  	 	break;
	   	  	 }
	   	  
	   }
	   cout<<c<<endl;
	}
}