Problem: problem

Given the lengths of $ 2 $ sides that are not the hypotenuse of a right triangle, $ A $ and $ B $. The side of length $ A $ overlaps the $ x $ axis, and the side of length $ B $ overlaps the $ y $ axis.

Do the following:

1. Rotate the triangle around the $ x $ axis.
2. Rotate the shape created by performing the operation $ 1 $ around the $ y $ axis.



Perform the operation $ 2 $ to find the volume of the created figure.



output

Output the volume of the figure. Also, output a line break at the end. Absolute or relative errors less than $ 0.000001 $ are allowed.

Example

Input

1 2


Output

33.510322

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
//type
typedef long long ll;
#define pii pair<int, int>;
#define vi vector<int>;
//x * y * 1.0 can cause overflow
//constant
#define inf (int)(1e9+7)
#define mod (ll)(1e9+7)
#define eps 1e-10
//omission
#define eb emplace_back
#define F first
#define S second
#define SZ(x) ((int)(x).size())
#define all(v) v.begin(), v.end()
#define ios cin.tie(0); ios::sync_with_stdio(false);
#define FIX(a) cout << fixed << setprecision(a)
//manip
#define chmax(a, b) a = max(a, b)
#define chmin(a, b) a = min(a, b)
#define UNIQUE(v) v.erase(unique(v.begin(), v.end(), v.end());
#define fill(x, y) memset(x, y, sizeof(x))
#define ceil(a, b) a / b + !!(a % b)
template<class T> T power(T a, T b)
{return b ? power(a * a % inf, b / 2) * (b % 2 ? a : 1) % inf : 1;}
#define LB(v, x) (int)(lower_bound(v.begin(), v.end(), x) - v.begin())
#define UB(v, x) (int)(upper_bound(v.begin(), v.end(), x) - v.begin())
template<typename T> T gcd(T a, T b) {if (b == 0) return a; return gcd(b, a % b);}
template<typename T> T lcm(T a, T b) {return a / gcd(max(a, b), min(a, b)) * b;} //先に割る!!
//loop
#define rep(i, n) for (int i = 0, i##_len = (n); i < i##_len; ++i)
#define rep3(i, st, n) for (int i = st; i < n; ++i)
//algorithm
//double pointer, l start, how many adds, can be 0 -> init r = l, sum = 0
//bfs, not x, y, i, j
//not !(i % 2), i % 2 == 0
//minimal required terms
//cast caution
//look constraints always

int main() {
    ios
    
    int a, b; cin >> a >> b;
    FIX(8);
    cout << 4.0 / 3 * M_PI * b * b * b << endl;
}

/*
 x方向に回転
 円錐ができる
 
 底辺は半径bの円
 
 a < bより、
 y方向に回転 -> 半径bの球
 
 
 */