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 int ll;
#define CC(t)            int t; cin>>t; while(t--)
#define FOR(i,a,b)      for (int i = a; i < b; i++)
int main()
{
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
  #ifndef ONLINE_JUDGE
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);  
 #endif
    ll t;
    cin>>t;
    while(t--)
    {
    	ll n;
    	cin>>n;
    	ll count = 0;
    	for(ll a=1;a<=n && (a*a+1)/2<=n;a++)
    	{
    		if((a*a-1)%2 != 0)
    			continue;
    		ll b = (a*a-1)/2;
    		ll c = b+1;
    		if(a<=b && b<=c && c<=n){
    			// cout<<a<<b<<c<<endl;
    			count++;
    		}
    	}
    	cout<<count<<endl;
    }
    return 0;
   		
}