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:
#if __has_include("debug.h")
    #include "debug.h"
#else  
#include <bits/stdc++.h>
using namespace std;
#define d(...) 2;
#endif

#define int long long int
#define ld long double

#define fast ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define endl "\n"      

#define all(c) (c).begin(),(c).end()
#define rall(c) (c).rbegin(),(c).rend()
#define ff first
#define ss second
#define eps 1e-10
#define PI acos(-1LL)

#define gcd(a,b)            __gcd((a),(b))
#define lcm(a,b)            ((a*b)/gcd(a,b))

void solve();

int32_t main()
{
    #ifdef DEBUG
	freopen("input.txt","r",stdin);
	freopen("output.txt","w",stdout);
    #endif 

    fast
    
    int t = 1;
	cin>>t;
	
	while(t--)
	    solve();
	return 0;  
}

void solve()
{
	int i,j;
	int n;
	cin>>n;

	int start=1,end=1e9+5;

	int ans=0;

	auto ok=[&](int x)
	{
          int y=2*x*x  + 2*x +1;
          return (y<=n);
	};

	while(start<=end)
	{
		int mid=start+(end-start)/2;
		if(ok(mid))
		{
			start=mid+1;
			ans=mid;
		}
		else
		{
			end=mid-1;
		}
	}

	cout<<ans<<endl;





}