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 int
#define scan(any)       for(auto &i:any) cin>>i;
#define print(any)      for(auto i:any) cout<<i<<" "; 
#define endl            '\n'
#define pb              push_back
#define vll             vector<ll> 
#define f(i,x,y)        for(i=x;i<y;i++)
#define INF             LLONG_MAX
#define s(x)            sort(x.begin(),x.end())
#define all(v)          v.begin(),v.end()
#define p2(n,x)         cout << fixed << setprecision(x) <<n<<endl; 
#define pll             pair<ll,ll> 
#define ff              first
#define ss              second
#define blt(x)          __builtin_popcount(x)

void solve()
{
    ll i,j,k,n;
    vll v;
    i=1;
    
    while(true)
    {

        ll a=i;
        ll b=(i*i)-1;
        b/=2;

        ll c=(a*a)+(b*b);

        c=sqrt(c);
        if(c>(1e9))
            break;

        if((a*a-1)%2==0 && (b>0) && b>=a)
            v.pb(c);

        i++;
    }
    ll q;
    cin>>q;
    while(q--)
    {
        cin>>n;
        ll u=upper_bound(all(v),n)-v.begin();
        u--;

        cout<<u+1<<endl;

    }
    //print(v);
    
}

int main()
{
    #ifndef ONLINE_JUDGE
        freopen("E:/Programming/input.txt", "r", stdin);
        freopen("E:/Programming/output.txt", "w", stdout);
    #endif

    ios::sync_with_stdio(0);
    cin.tie(0);
    
    ll t = 1;
    while(t--) {
        solve();    
    }
    return 0;
}