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>
using namespace std;
typedef long long ll;
typedef long double ld;
#define int ll
#define f first
#define s second
#define ins insert
#define er erase
#define pb push_back
#define ppb pop_back
#define pf push_front
#define ppf pop_front
#define INF (ll)1e18

int bl(int x){return 2*x*x+2*x+1;}

int bs(int k){
    int l=-1e6,r=1e6;
    while(r>l+1){
        int c=(l+r)/2;
        if(bl(c)>k)r=c;
        else l=c;
    }
    return l;
}

signed main(){
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    int t;
    cin>>t;
    while(t--){
        int n;
        cin>>n;
        int a=bs(n);
        cout<<a<<"\n";
    }
}

/*
a = m^2-n^2
b = 2mn
c = m^2+n^2

m^2 = (a+c)/2;
n^2 = (c-a)/2;

int this task
c = a^2 - b
a <= b <= c
*/