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>
#define rep(i,a,b) for(int i=a;i<b;i++)
#define rrep(i,a,b) for(int i=a;i>=b;i--)
#define fore(i,a) for(auto &i:a)
#define all(x) (x).begin(),(x).end()
#pragma GCC optimize ("-O3")
using namespace std; void _main(); int main() { cin.tie(0); ios::sync_with_stdio(false); _main(); }
typedef long long ll; const int inf = INT_MAX / 2; const ll infl = 1LL << 60;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a = b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a = b; return 1; } return 0; }
//---------------------------------------------------------------------------------------------------
/*---------------------------------------------------------------------------------------------------
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　　　　 （ ´_ゝ`）　/　 ⌒i     
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　 　　　＼/＿＿＿＿/　（u　⊃  
---------------------------------------------------------------------------------------------------*/



const double PI = 2 * acos(0.0);
double A, B;
//---------------------------------------------------------------------------------------------------
void _main() {
    cin >> A >> B;

    double r = max(A, B);
    double ans = 4.0 * PI * r * r * r / 3;

    printf("%.10f\n", ans);
}