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:
///bismillahir rahmanir rahim

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define dl double
#define st string
#define ch char
#define fl bool flag=false
#define cn(n) cin>>n
#define cll(n) long long n; cin>>n
#define cdl(n) double n; cin>>n
#define cst(s) string s; cin>>s
#define ct(n) cout<<n
#define w(n) while(n>0)
#define elif else if
#define wh(t) long long t; cin>>t; while(t--)
#define a(m, n) long long a[n]; for(int i=m; i<n; i++) cin>>a[i]
#define cf(m, i, n) for(int i=m; i<n; i++) cout<<a[i]<<" ";
#define c(m, n) long long c[n+1]; for(int i=m; i<=n; i++) c[i]=0
#define f(m, i, n) for(int i=m; i<n; i++)
#define fr(m, i, n) for(int i=m-1; i>=n; i--)
#define srt(a) sort(a, a+n)
#define srrt(a) sort(a, a+n, greater<int>())
#define e(n) if(n%2==0);
#define o(n) if(n%2==1);
#define yes cout<<"YES"<<endl
#define no cout<<"NO"<<endl
#define bye cout<<endl

int main()
{
    wh(t){
        cll(n);
        ll m=sqrt(2*n-1);
        if(m%2==0) m-=1;
        ct((m-1)/2);
        bye;
    }
}