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>
#include <cmath>
using namespace std;

typedef long long ll;

typedef vector<int> vi;
typedef vector<bool> vb;
typedef vector<vi> vvi;
typedef vector<long long> vll;
typedef vector<pair<int, int>> vpii;
typedef vector<pair<ll, ll>> vpll;

typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

typedef queue<int> qi;
typedef queue<ll> qll;
typedef deque<int> dqi;
typedef deque<ll> dqll;

#define pb(x) push_back(x)
#define pob() pop_back()

#define r(i, s, n) for(ll i = s; i < n; i++)
#define rr(i, e, s) for(ll i = e; i >= s; i--)

#define MAX 1e9+7
#define sf bool f=true
#define fi first
#define se second
#define YES cout<<"YES"<<endl;
#define NO cout<<"NO"<<endl;

// a*a = 2*b+1
// For all odd integers `a`, there exists only 1 b
// c = b+1

void solve(){

    int n;
    cin>>n;
    int ans = 0;

    for(int i = 3; i*i <= 2*n-1; i+=2){
        ans++;
    }

    cout<<ans<<endl;

}

int main(){

    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);


    int t;
    cin>>t;
    while(t--) solve();



    return 0;
}