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>

using namespace std;

#define DBG cerr << '!' << endl;
#define REP(i,n) for(ll (i) = (0);(i) < (n);++i)
#define rep(i,s,g) for(ll (i) = (s);(i) < (g);++i)
#define rrep(i,s,g) for(ll (i) = (s);i >= (g);--(i))
#define PB push_back
#define MP make_pair
#define FI first
#define SE second
#define SHOW1d(v,n) {for(int W = 0;W < (n);W++)cerr << v[W] << ' ';cerr << endl << endl;}
#define SHOW2d(v,i,j) {for(int aaa = 0;aaa < i;aaa++){for(int bbb = 0;bbb < j;bbb++)cerr << v[aaa][bbb] << ' ';cerr << endl;}cerr << endl;}
#define ALL(v) v.begin(),v.end()
#define Decimal fixed<<setprecision(10)
#define INF 1000000000
#define MOD 1000000007

typedef long long ll;
typedef pair<ll,ll> P;

double dp[555][555];

int main()
{
	int n,m;cin >> n >> m;
	vector<double> v(n+1);
	REP(i,n)cin >> v[i+1];
	
	for(int k = 0;k < m;k++){
		for(int i = k;i < n;i++){
			if(!(i == 0 && k == 0)&& dp[i][k] == 0)continue;
			double sum = 0;
			for(int j = 1;i+j <= n;j++){
				sum += v[i+j];	
				dp[i+j][k+1] = max(dp[i+j][k+1],dp[i][k] + (sum/j));
			}
		}
	}
	
	//SHOW2d(dp,n+1,m+1);
	
	cout << Decimal << dp[n][m] << endl;

	
	return 0;
}