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:
//Everything Is Finished !!
//So Near Yet So Far !! 
#include<bits/stdc++.h>
using namespace std;

#define fast ios_base::sync_with_stdio(0);cin.tie(NULL);cout.tie(NULL)
#define ll long long
#define ld long double
#define pb push_back
#define fe first
#define se second
#define nl "\n"
#define pp pair < ll , ll > 
#define sz(x) (ll)x.size()
#define st(x) sort(x.begin(),x.end())
#define rst(x) sort(x.rbegin(), x.rend())
#define all(x) x.begin(),x.end()
long double pi = 3.14159265358979323;
 
const double EPS = 1e-12;
const int N = 1e6 + 5;
//Just A Gentle Reminder about MOD
const int mod = 1e9 + 7;
 
 
bool check(ll val , ll n){
	return val*val <= 2*n - 1;
} 
 
int main()
{
    fast;
    //pre();
    ll t;
    cin >> t ;
    while(t--){
    	ll n;
    	cin >> n ;
    	ll low = 1 , high = n ;
    	while(low <= high){
    		ll mid = (low + (high - low )/2);
    		if(check(mid , n)){
    			low = mid + 1;
    		}
    		else{
    			high = mid - 1;
    		}
    	}
    	//cout << low << " ";
    	cout << (low - 2)/2 << "\n";
    }
    





















    
    return 0;
//It's not end it yet
}