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:
/* >>>>>Shwetal<<<<< */
#include <bits/stdc++.h>
#define int long long
#define f(i, begin, end) for (__typeof(end) i = (begin) - ((begin) > (end)); i != (end) - ((begin) > (end)); i += 1 - 2 * ((begin) > (end)))
#define inputArray(arr) f(i, 0, arr.size()) cin >> arr[i]
#define printArray(arr) f(i, 0, arr.size()) cout << arr[i] << ' '; cout << endl
#define endl '\n'
typedef long long ll;
const int mod = 1e9+7;
using namespace std;

void solve() {
    int n;
    cin >> n;
    int l = 1;
    int r = n/2;
    int ans = 0;
    while(l <= r){
        int m  = l + (r - l)/2;
        int a = 2*m + 1;
        int b = (a*a - 1)/2;
        int c = a*a - b;
        if(c <= n){
            ans = m;
            l = m+1;
        }else{
            r = m - 1;
        }
    }
    cout << ans << endl;
}

int32_t main(){
    // freopen("input.txt","r",stdin); freopen("output.txt","w",stdout);
    ios::sync_with_stdio(0);
    cin.tie(0);
    int T = 1;
    cin >> T;
    while(T--){
        solve();
        // if(solve())
            // cout << "YES" << endl;
        // else
            // cout << "NO" << endl;
    }
    return 0;
}