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>
#define ll long long
#define pb push_back
#define mp make_pair
using namespace std;
const ll infl = 1LL << 60;
ll mod= 998244353;


 
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin>>t;
  while (t--) {
      ll n;
      cin>>n;
      if(n<5){
        cout<<0<<endl;
        continue;
      }
      //ll k = 3;
      ll count=0;
      for(ll k = 3; (((k*k)+1)/2)<=n; k+=2){
         count++;
      }
      cout<<count<<endl;




      
  }
  return 0;
}
//DEBUG
//memset doesnt work for values other than 0 or -1
//use ll
//check base case for n=1
//out of bounds(n==1 !!!)
//int ovetflow(a-b+mod !!!)
//use & for containers to modify in function calling