#include<bits/stdc++.h>
#define db long double
#define For(i,a,b) for(int i=(a);i<=(b);i++)
using namespace std;
const int N=1e5+5;
typedef long long ll;
const db PI=acos(-1);
struct node{
	ll x,y;
	node operator +(node B){
		return {(x+B.x),(y+B.y)};
	}
	friend node operator *(int k,node A){
		return {k*A.x,k*A.y};
	}
	inline void operator +=(node B){
		*this=*this+B;
	} 
};
int inv2;
int n,m;
inline node gt(int x){
	if(x==0)return {0,0};
	int k=floor(2.0*m/PI);
	int las=2*m-1-k*2;
	int p1=(1ll*k*(k+1)*x);
	return {p1,2ll*m*x*las};
}
node solve(int x){
	node ans=gt(x);
	ll res=1ll*x*(x-1)/2;
	res=res*4*m*m;
	res+=2ll*x*m;
	ans.y+=res;
	return ans;
}
int main(){
//	freopen("a6.in","r",stdin);
//	freopen("a.out","w",stdout);
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin>>n>>m;
	if(m==1){
		ll ans=0;
		for(int i=-n;i<=n;i++){
			if(i==0)continue;
			for(int j=i+1;j<=n;j++){
				if(j==0)continue;
				ans+=j-i;
			}
		}
		cout<<fixed<<setprecision(6)<<ans;
		return 0;
	}
	node ans={0,0};
	for(int i=1;i<=n;i++)ans+=solve(i);
	for(int i=0;i<n;i++)ans+=2ll*(n-i)*gt(i);
	cout<<fixed<<setprecision(6)<<PI*ans.x+ans.y<<"\n";
	return 0;
}