#define _USE_MATH_DEFINES
#include "bits/stdc++.h"
using namespace std;
#define FOR(i,j,k) for(int (i)=(j);(i)<(int)(k);++(i))
#define rep(i,j) FOR(i,0,j)
#define each(x,y) for(auto &(x):(y))
#define mp make_pair
#define mt make_tuple
#define all(x) (x).begin(),(x).end()
#define debug(x) cout<<#x<<": "<<(x)<<endl
#define smax(x,y) (x)=max((x),(y))
#define smin(x,y) (x)=min((x),(y))
#define MEM(x,y) memset((x),(y),sizeof (x))
#define sz(x) (int)(x).size()
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<ll> vll;
 
inline double heron(double a, double b, double c){
    double s = (a + b + c) / 2;
    return sqrt(s * (s-a) * (s-b) * (s-c));
}
 
int N;
ll K;
double d[200001];
 
inline double g(int i, int j, int k) {
    double a = d[j - i], b = d[k - j], c = d[k - i];
    return heron(a, b, c);
}
 
int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    while(cin >> N >> K) {
        rep(i, N) {
            double w = 2 * M_PI * i / N;
            double x = cos(w), y = sin(w);
            d[i] = hypot(x - 1, y);
        }
 
        double l = 0, r = 2, m;
        rep(i_, 35) {
            m = (l + r) / 2;
            ll z = 0;
            for(int i = 1; i < N - 1; ++i) {
                int j = (N + i) / 2;
                int L = i + 1, R = j + 1, M;
                if(g(0, i, i+1) > m) continue;
                while(R - L > 1) {
                    M = (L + R) / 2;
                    (g(0, i, M) <= m ? L : R) = M;
                }
                if(L == j)z += (ll)(N - i - 1)*i;
                else z += 2 * (L - i)*i;
            }
            (z >= K ? r : l) = m;
        }
        cout << setprecision(20) << l << endl;
    }
}
### Prompt

Your task is to create a Cpp solution to the following problem:

Input

The input is given from Standard Input in the following format:


> $N \ K$

Output

* Please output the $K$-th triangle area.
* Print a floating number denoting the answer. The relative or absolute error of your answer should not be higher than $10^{−9}$.



Constraints

* $3 \le N \le 200,000$
* $1 \le K \le \frac{N(N-1)(N-2)}{6}$



Subtasks

Subtask 1 [ $160$ points ]


* $N \le 100$

Subtask 2 [ $240$ points ]


* $N \le 1000$

Subtask 3 [ $450$ points ]


* $N \le 200,000$

Output

* Please output the $K$-th triangle area.
* Print a floating number denoting the answer. The relative or absolute error of your answer should not be higher than $10^{−9}$.



Constraints

* $3 \le N \le 200,000$
* $1 \le K \le \frac{N(N-1)(N-2)}{6}$



Subtasks

Subtask 1 [ $160$ points ]


* $N \le 100$

Subtask 2 [ $240$ points ]


* $N \le 1000$

Subtask 3 [ $450$ points ]


* $N \le 200,000$

Input

The input is given from Standard Input in the following format:


> $N \ K$

Examples

Input

4 3


Output

1.0000000000000


Input

6 9


Output

0.86602540378


Input

12 220


Output

1.29903810568

### Response

```cpp
#define _USE_MATH_DEFINES
#include "bits/stdc++.h"
using namespace std;
#define FOR(i,j,k) for(int (i)=(j);(i)<(int)(k);++(i))
#define rep(i,j) FOR(i,0,j)
#define each(x,y) for(auto &(x):(y))
#define mp make_pair
#define mt make_tuple
#define all(x) (x).begin(),(x).end()
#define debug(x) cout<<#x<<": "<<(x)<<endl
#define smax(x,y) (x)=max((x),(y))
#define smin(x,y) (x)=min((x),(y))
#define MEM(x,y) memset((x),(y),sizeof (x))
#define sz(x) (int)(x).size()
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<ll> vll;
 
inline double heron(double a, double b, double c){
    double s = (a + b + c) / 2;
    return sqrt(s * (s-a) * (s-b) * (s-c));
}
 
int N;
ll K;
double d[200001];
 
inline double g(int i, int j, int k) {
    double a = d[j - i], b = d[k - j], c = d[k - i];
    return heron(a, b, c);
}
 
int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    while(cin >> N >> K) {
        rep(i, N) {
            double w = 2 * M_PI * i / N;
            double x = cos(w), y = sin(w);
            d[i] = hypot(x - 1, y);
        }
 
        double l = 0, r = 2, m;
        rep(i_, 35) {
            m = (l + r) / 2;
            ll z = 0;
            for(int i = 1; i < N - 1; ++i) {
                int j = (N + i) / 2;
                int L = i + 1, R = j + 1, M;
                if(g(0, i, i+1) > m) continue;
                while(R - L > 1) {
                    M = (L + R) / 2;
                    (g(0, i, M) <= m ? L : R) = M;
                }
                if(L == j)z += (ll)(N - i - 1)*i;
                else z += 2 * (L - i)*i;
            }
            (z >= K ? r : l) = m;
        }
        cout << setprecision(20) << l << endl;
    }
}
```