#include <iostream>
#include <algorithm>
#include <vector>
#include <cmath>
#include <iomanip>
using namespace std;
template<class T> inline void YES(T condition){ if(condition) cout << "YES" << endl; else cout << "NO" << endl; }
template<class T> inline void Yes(T condition){ if(condition) cout << "Yes" << endl; else cout << "No" << endl; }
template<class T> inline void POSS(T condition){ if(condition) cout << "POSSIBLE" << endl; else cout << "IMPOSSIBLE" << endl; }
template<class T> inline void Poss(T condition){ if(condition) cout << "Possible" << endl; else cout << "Impossible" << endl; }
template<class T> inline void First(T condition){ if(condition) cout << "First" << endl; else cout << "Second" << endl; }
template<class T = string, class U = char>int character_count(T text, U character){ int ans = 0; for(U i: text){ ans += (i == character); } return ans; }
long power(long base, long exponent, long module){ if(exponent % 2){ return power(base, exponent - 1, module) * base % module; }else if(exponent){ long root_ans = power(base, exponent / 2, module); return root_ans * root_ans % module; }else{ return 1; }}
struct position{ int y, x; }; position move_pattern[4] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}}; // double euclidean(position first, position second){ return sqrt((second.x - first.x) * (second.x - first.x) + (second.y - first.y) * (second.y - first.y)); }
template<class T, class U> string to_string(pair<T, U> x){ return to_string(x.first) + "," + to_string(x.second); }
template<class itr> void array_output(itr start, itr goal){ string ans; for(auto i = start; i != goal; i++){ ans += to_string(*i) + " "; } ans.pop_back(); cout << ans << endl; }
template<class itr> void cins(itr start, itr goal){ for(auto i = start; i != goal; i++){ cin >> (*i); } }
template<class T> T gcd(T a, T b){ if(a && b){ return gcd(min(a, b), max(a, b) % min(a, b)); }else{ return a; }} template<class T> T lcm(T a, T b){ return a / gcd(a, b) * b; }
#define mod long(1e9 + 7)
#define all(x) (x).begin(), (x).end()
#define bitcount(n) __builtin_popcountl(long(n))
#define fcout cout << fixed << setprecision(20)
#define highest(x) (63 - __builtin_clzl(x))

const double pi = 3.1415926535;

int main(){
    long N, K;
    cin >> N >> K;
    double left = 0, right = 100;
    for(int i = 0; i < 100; i++){
        double middle = (left + right) / 2;
        long less_middle = 0;
        for(long j = 1; j < N; j++){
            
            long left2 = j - 1, right2 = (N - j) / 2 + 1; // [left2, right2)
            if(left2 > right2){
                continue;
            }
            while(left2 + 1 < right2){
                long middle2 = (left2 + right2) / 2;
                long l = N - j - middle2;
                double area = 2 * sin(pi / N * j) * sin(pi / N * middle2) * sin(pi / N * l);
                if(area >= middle){
                    right2 = middle2;
                }else{
                    left2 = middle2;
                }
            }
            
            long max_k = left2 + 1;
            if(j == max_k){
                continue;
            }
            less_middle += (max_k - j) * N * 2;
            if(N % 3 == 0 && j == N / 3 && max_k > j){
                less_middle += N / 3 - N * 2;
            }else{
                if(max_k > j){
                    less_middle += N - N * 2;
                }
                if((N - j) % 2 == 0 && max_k > (N - j) / 2){
                    less_middle += N - N * 2;
                }
            }
        }
        if(less_middle < K){
            left = middle;
        }else{
            right = middle;
        }
    }
    fcout << left << endl;
}

### Prompt

Please formulate 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
#include <iostream>
#include <algorithm>
#include <vector>
#include <cmath>
#include <iomanip>
using namespace std;
template<class T> inline void YES(T condition){ if(condition) cout << "YES" << endl; else cout << "NO" << endl; }
template<class T> inline void Yes(T condition){ if(condition) cout << "Yes" << endl; else cout << "No" << endl; }
template<class T> inline void POSS(T condition){ if(condition) cout << "POSSIBLE" << endl; else cout << "IMPOSSIBLE" << endl; }
template<class T> inline void Poss(T condition){ if(condition) cout << "Possible" << endl; else cout << "Impossible" << endl; }
template<class T> inline void First(T condition){ if(condition) cout << "First" << endl; else cout << "Second" << endl; }
template<class T = string, class U = char>int character_count(T text, U character){ int ans = 0; for(U i: text){ ans += (i == character); } return ans; }
long power(long base, long exponent, long module){ if(exponent % 2){ return power(base, exponent - 1, module) * base % module; }else if(exponent){ long root_ans = power(base, exponent / 2, module); return root_ans * root_ans % module; }else{ return 1; }}
struct position{ int y, x; }; position move_pattern[4] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}}; // double euclidean(position first, position second){ return sqrt((second.x - first.x) * (second.x - first.x) + (second.y - first.y) * (second.y - first.y)); }
template<class T, class U> string to_string(pair<T, U> x){ return to_string(x.first) + "," + to_string(x.second); }
template<class itr> void array_output(itr start, itr goal){ string ans; for(auto i = start; i != goal; i++){ ans += to_string(*i) + " "; } ans.pop_back(); cout << ans << endl; }
template<class itr> void cins(itr start, itr goal){ for(auto i = start; i != goal; i++){ cin >> (*i); } }
template<class T> T gcd(T a, T b){ if(a && b){ return gcd(min(a, b), max(a, b) % min(a, b)); }else{ return a; }} template<class T> T lcm(T a, T b){ return a / gcd(a, b) * b; }
#define mod long(1e9 + 7)
#define all(x) (x).begin(), (x).end()
#define bitcount(n) __builtin_popcountl(long(n))
#define fcout cout << fixed << setprecision(20)
#define highest(x) (63 - __builtin_clzl(x))

const double pi = 3.1415926535;

int main(){
    long N, K;
    cin >> N >> K;
    double left = 0, right = 100;
    for(int i = 0; i < 100; i++){
        double middle = (left + right) / 2;
        long less_middle = 0;
        for(long j = 1; j < N; j++){
            
            long left2 = j - 1, right2 = (N - j) / 2 + 1; // [left2, right2)
            if(left2 > right2){
                continue;
            }
            while(left2 + 1 < right2){
                long middle2 = (left2 + right2) / 2;
                long l = N - j - middle2;
                double area = 2 * sin(pi / N * j) * sin(pi / N * middle2) * sin(pi / N * l);
                if(area >= middle){
                    right2 = middle2;
                }else{
                    left2 = middle2;
                }
            }
            
            long max_k = left2 + 1;
            if(j == max_k){
                continue;
            }
            less_middle += (max_k - j) * N * 2;
            if(N % 3 == 0 && j == N / 3 && max_k > j){
                less_middle += N / 3 - N * 2;
            }else{
                if(max_k > j){
                    less_middle += N - N * 2;
                }
                if((N - j) % 2 == 0 && max_k > (N - j) / 2){
                    less_middle += N - N * 2;
                }
            }
        }
        if(less_middle < K){
            left = middle;
        }else{
            right = middle;
        }
    }
    fcout << left << endl;
}

```