Problem: problem

AOR Ika got a water tank with a size of $ 1 $ in length and $ N $ in width. The aquarium is tall enough to hold water. The aquarium has $ N-1 $ partitions and is evenly spaced into $ N $ compartments. When water was poured here, the height of the water in each section became $ a_i $.

AOR Ika decided to remove some dividers and reduce the number of compartments to $ M $ or less. When you have removed the dividers, find the maximum sum of the water heights in each compartment.

The thickness of the partition can be ignored.



input

$ N \ M $
$ a_1 \ cdots a_N $

output

Output the maximum value of the total water height of each section in one line. Also, output a line break at the end. It is acceptable if the relative or absolute error is less than or equal to $ 10 ^ {-6} $.

Example

Input

5 3
9 1 2 3 9


Output

20.000000

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define rep1(i,n) for(int i=1;i<=(int)(n);i++)
#define all(c) c.begin(),c.end()
#define pb push_back
#define fs first
#define sc second
#define show(x) cout << #x << " = " << x << endl
#define chmin(x,y) x=min(x,y)
#define chmax(x,y) x=max(x,y)
using namespace std;
template<class S,class T> ostream& operator<<(ostream& o,const pair<S,T> &p){return o<<"("<<p.fs<<","<<p.sc<<")";}
template<class T> ostream& operator<<(ostream& o,const vector<T> &vc){o<<"sz = "<<vc.size()<<endl<<"[";for(const T& v:vc) o<<v<<",";o<<"]";return o;}

int N,M;
double h[500];
double dp[510][510];

int main(){
	cin>>N>>M;
	rep(i,N) cin>>h[i];
	rep(i,N+1) rep(j,M+1) dp[i][j] = -1e50;
	dp[0][0] = 0;
	rep(i,N){
		rep(j,M) if(dp[i][j]>=0){
			double acc = 0;
			for(int k=i+1;k<=N;k++){
				acc += h[k-1];
				chmax(dp[k][j+1],dp[i][j] + acc/(k-i));
			}
		}
	}
	double ans = 0;
	rep1(j,M) chmax(ans,dp[N][j]);
	printf("%.12f\n",ans);
}