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

using namespace std;

vector<ll> nums;

int binse(int l, int r, int n){
  if(r-l==1)return l;
  else if(r==l)return l;
  else{
    int mid = (l+r)/2;
    int mid1 = mid+1;
    if(n>nums[l] && n<nums[mid]){
      return binse(l,mid,n);
    }else if(n>nums[mid1] && n<nums[r]){
      return binse(mid1,r,n);
    }else if(n>nums[mid] && n<nums[mid1]){
      return mid;
    }else if(n==nums[l]){
      return l;
    }else if(n==nums[mid]){
      return mid;
    }else if(n==nums[mid1]){
      return mid1;
    }else{
      return r;
    }
  }
}



int main(){
  ll t;
  ll n, i, j ,k , l;
  for(i = 1;((i*i)+1)<=(2000000000);i+=2){
    nums.push_back(((i*i)+1)/2);
  }
  cin>>t;
  n = nums.size();
  while(t--){
    cin>>l;
    i = binse(0,n-1,l);
    cout<<i<<endl;
  }
}