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;

/*<DEBUG>*/
#define tem template <typename 
#define can_shift(_X_, ...) enable_if_t<sizeof test<_X_>(0) __VA_ARGS__ 8, debug&> operator<<(T i)
#define _op debug& operator<<
tem C > auto test(C *x) -> decltype(cerr << *x, 0LL);
tem C > char test(...);
tem C > struct itr{C begin, end; };
tem C > itr<C> get_range(C b, C e) { return itr<C>{b, e}; }
struct debug{
#ifdef _LOCAL
	~debug(){ cerr << endl; }
	tem T > can_shift(T, ==){ cerr << boolalpha << i; return *this; }
	tem T> can_shift(T, !=){ return *this << get_range(begin(i), end(i)); }
	tem T, typename U > _op (pair<T, U> i){ 
		return *this << "< " << i.first << " , " << i.second << " >"; }
	tem T> _op (itr<T> i){
		*this <<  "{ ";
		for(auto it = i.begin; it != i.end; it++){
			*this << " , " + (it==i.begin?2:0) << *it;
		}
		return *this << " }";
	}
#else
tem T> _op (const T&) { return *this; }
#endif 
};

string _ARR_(int* arr, int sz){
	string ret = "{ " + to_string(arr[0]); 
	for(int i = 1; i < sz; i++) ret += " , " +  to_string(arr[i]);
	ret += " }"; return ret;
}

#define exp(...) " [ " << #__VA_ARGS__ << " : " << (__VA_ARGS__) << " ]"
/*</DEBUG>*/

typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;
typedef pair<int, int> pii;
//mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());

#define pb push_back
#define FAST ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define TC int __TC__; cin >> __TC__; while(__TC__--)
#define ar array

const int INF = 1e9 + 7;

int main(void)
{
	FAST;
	vector<ll> mx;
	for(ll i = 1; i*i + (i+1)*(i+1) <= INF; ++i){
		mx.pb(i*i + (i+1) * (i+1));
	}

	TC{
		int n; cin >> n;
		int pos = (lower_bound(mx.begin(), mx.end(), n) - mx.begin());
		if(mx[pos] == n) ++pos;
		cout << pos << '\n';
	}

	return 0;
}