我们这道题考虑用总答案除以方案数，然后就会发现相当于对于每一个位置被覆盖成 $0$ 的概率之和

首先我们可以通过一些简单的方法求出来总线段个数（$O(1)$ 方法），即 $\sum_{i=n-B+1}^{n-A+1} i$

然后我们对于每一个点类似地求被覆盖的线段个数，然后统计答案即可

#include<iostream>
#include<iomanip>
#include<cstdio>
#include<cmath>
#define int long long
using namespace std;
bool MLE_Test_start;
const int N=1e6+10;
int T=1,n,m,A,B,sum1,now[N];
long double ret=0,ans[N];
inline int reads(){
	char c=getchar();
	int x=0,f=1;
	while(!isdigit(c)){
		if(c=='-') f=-1;
		c=getchar();
	}
	while(isdigit(c)){
		x=(x<<3)+(x<<1)+(c^48);
		c=getchar();
	}
	return x*f;
}
inline void files(){
//	freopen("std.in","r",stdin);
	freopen("std.out","w",stdout);
}
inline void clr(){
	//	Don't Forget!
	
}
int asks(int l,int r){return (l+r)*(r-l+1)/2;}
bool MLE_Test_end;
signed main(){
	//	printf("%lf\n",(&MLE_Test_end-&MLE_Test_start-1)/1024.0/1024.0);
//		files();
	//	T=reads();
	while(T--){
		clr();
		n=reads(),m=reads(),A=reads(),B=reads();
		sum1=asks(n-B+1,n-A+1);	
//		cout<<sum1<<"\n";
		for(int i=1;i<=n;i++){
			int lft=min(i-1,n-i),rght=max(i-1,n-i);
//			now[i]=asks(min(max(rght-A+1),max(0,lft-B+1)),max(max(0,lft-B+1),max(0,rght-A+1)));
			now[i]=asks(max(0ll,rght-B+1),max(0ll,rght-A+1))+asks(max(0ll,lft-B+1),max(0ll,lft-A+1));
//			cout<<i<<":"<<now[i]<<"\n";
//			now[i]=sum1-now[i];
			ans[i]=1-1.0*pow((long double)now[i]/sum1,m);
			ret+=ans[i]*1.0;
//			cout<<now[i]<<" "<<ans[i]<<" "<<(long double)now[i]/sum1<<" "<<lft<<" "<<rght<<" "<<"\n";
		}
		cout<<fixed<<setprecision(3)<<ret<<"\n";
	}
	return 0;
}