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<iostream>
#include<vector>
#include<algorithm>
#include<set>
#include<map>
#include<unordered_map>
#include<cmath>
#include<deque>
#include<queue>
#include<stack>
#include<random>
#include<time.h>
using namespace std;
#define int long long
typedef pair<int, int> pii;
const int mod = 1e9 + 7;
const int INF = 1e18;
bool ok(int a, int n) {
    return (1 <= a && a <= n);
}
void solve() {
    int n;
    cin >> n;
    int ans = 0;
    for (int a = 1;; a++) {
        if (a * a > 2 * n) break;
        if ((a * a - 1) % 2) continue;
        int b = (a * a - 1) / 2;
        int c = b + 1;
        if (ok(b, n) && ok(c, n)) ans++;
    }
    cout << ans << '\n';
}
int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int tt = 1;
    cin >> tt;
    while (tt--) {
        solve();
        cout << '\n';
    }
}