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
#include<bits/stdc++.h>
using namespace std;
using Int = long long;

template<typename T>
vector<T> make_v(size_t a){return vector<T>(a);}
template<typename T>
vector<vector<T> > make_v(size_t a,size_t b){
  return vector<vector<T> >(a,make_v<T>(b));
}
template<typename T>
vector<vector<vector<T> > > make_v(size_t a,size_t b,size_t c){
  return vector<vector<vector<T> > > (a,make_v<T>(b,c));
}

template<typename T,typename V>
typename enable_if<is_class<T>::value==0>::type
fill_v(T &t,const V &v){t=v;}

template<typename T,typename V>
typename enable_if<is_class<T>::value!=0>::type
fill_v(T &t,const V &v){
  for(auto &e:t) fill_v(e,v);
}


template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}


struct Precision{
  Precision(){
    cout<<fixed<<setprecision(12);
  }
}precision_beet;

//INSERT ABOVE HERE
signed main(){
  Int n,m;
  cin>>n>>m;
  vector<double> a(n);
  for(Int i=0;i<n;i++) cin>>a[i];

  const double INF = 1e18;
  auto dp=make_v<double>(n+1,m+1);
  fill_v(dp,-INF);
  dp[0][0]=0;
  for(Int i=0;i<n;i++){
    for(Int j=0;j<m;j++){
      double sum=0;
      for(Int k=1;i+k<=n;k++){
	sum+=a[i+k-1];
	chmax(dp[i+k][j+1],dp[i][j]+sum/k);
      }
    }
  }
  
  cout<<dp[n][m]<<endl;
  return 0;
}

