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>
typedef long long ll;
using namespace std;
#define fastio ios_base::sync_with_stdio(false);cin.tie(0)
#define all(x) x.begin(),x.end()
#define pb push_back
#define pf push_front
#define ff first
#define ss second
#define vll vector<ll>
#define sll set<ll>
#define pll pair<ll, ll>
#define vpll vector<pll>
#define endl "\n"
const ll maxn = 1e6+2;
const ll MOD=1000000007;
/*******************************************************************************/
ll mod_sum() { return 0LL; }
template < typename T, typename... Args >
T mod_sum(T a, Args... args) { return ((a + mod_sum(args...))%MOD + MOD)%MOD; }
ll mod_prod() { return 1LL; }
template< typename T, typename... Args >
T mod_prod(T a, Args... args) { return (a * mod_prod(args...))%MOD; }
/*******************************************************************************/

int main()
{
    fastio; //remove for interactive
    ll i, j, k, n, t;

    cin>>t;
    while (t--)
    {
        cin>>n;

        ll ans = 0;
        for (ll a = 3; (a*a+1)/2 <= n; a+=2)
            ans++;
        

        cout<<ans<<endl;
    }
    

    return 0;
}