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:
/**
 *    author:  Mohamad Milhem
**/

#include <bits/stdc++.h>
#include <stdio.h>

using namespace std;
typedef long long ll;

#define FAST ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define mp make_pair
#define pb push_back
#define lp(i,s,f) for(ll i = s; i < ll(f); i++)
#define inF freopen("input.in", "r", stdin);
#define outF freopen("output.in", "w", stdout);
#define MOD ll(1000000007)
#define debug(x) cout << '[' << #x << " is: " << x << "] " <<endl;
#define decimalpoint cout << std::fixed << setprecision(5)



int m[105][105];


void solve(){


	int n;
	cin >> n;


	int low = 0;
	int high = n;


	while (low <= high){
		int mid = (low + high) / 2;

		long long number = mid * 2 + 1;


		if ((number*number+1)/2 > n){
			high = mid - 1;
		}

		else {
			low = mid + 1;
		}


	}

	cout << low - 1 << endl;



}



int main()
{
    FAST;
    int t;
    cin >> t;

    while(t--){
        solve();
    }
    return 0;
}