#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

const int lim = 2000000007;

vector <int> seq;
int T;
int n;

int main()
{
	for (ll i = 3; i * i <= lim; i += 2)
		seq.push_back((i * i + 1) / 2);
	scanf("%d", &T);
	while (T--) {
		scanf("%d", &n);
		int ind = upper_bound(seq.begin(), seq.end(), n) - seq.begin();
		printf("%d\n", ind);
	}
    return 0;
}

### Prompt

Your challenge is to write a Cpp solution to the following 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

const int lim = 2000000007;

vector <int> seq;
int T;
int n;

int main()
{
	for (ll i = 3; i * i <= lim; i += 2)
		seq.push_back((i * i + 1) / 2);
	scanf("%d", &T);
	while (T--) {
		scanf("%d", &n);
		int ind = upper_bound(seq.begin(), seq.end(), n) - seq.begin();
		printf("%d\n", ind);
	}
    return 0;
}

```