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>
using namespace std;
#define ll long long int	
#define fr first
#define sc second
#define pb push_back
#define sz(c) c.size()
#define pu push
#define po pop
#define ins insert
#define srt(v)  sort(v.begin(),v.end())
#define rep(i, a, b) for (int i=a; i<b; i++)
#define rep2(i, a, b) for (int i=a; i>=b; i--)
#define mem0(a) memset(a,0,sizeof(a))
#define mem1(a) memset(a,1,sizeof(a))

int main(){
	int t;
	cin>>t;
	while(t--){
		ll n;
		cin>>n;
		ll le=1;
		ll re=1e5;
		ll mx=0;
		while(le<=re){
			ll m=(le+re)/2;
			ll te=(m*(m+1))/2;
			if(4*te+1<=n){
				mx=max(mx,m);
				le=m+1;}
			else
			re=m-1;}
		cout<<mx<<endl;
	}
}