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;
#define ll long long
#define pb push_back
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
 
ll int gcd(ll int a,ll int b){if(!b)return a;return gcd(b, a % b);}
ll int lcm(ll int a,ll int b){ return (a*b)/gcd(a,b);}
template<typename T> T pow(T a,T b, long long m){T ans=1; while(b>0){ if(b%2==1) ans=(ans*a)%m; b/=2; a=(a*a)%m; } return ans%m; }



  int main(){
    IOS;
  ll int tt;  cin>>tt; while(tt--){
   ll int n,cnt=0;
   cin>>n;
    if(n<=4) {printf("0\n"); continue;}
   for(int i=3;i<=1000000000;i+=2){
      if((((i*i)-1)/2)+1<=n) cnt++;
      else break;
    //  printf("%lld ",s);
      
   }
   printf("%lld\n",cnt);
 }
 }