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 <cstdio>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <cfloat>
#include <cstring>
#include <map>
#include <utility>
#include <set>
#include <iostream>
#include <memory>
#include <string>
#include <vector>
#include <list>
#include <algorithm>
#include <functional>
#include <sstream>
#include <complex>
#include <stack>
#include <queue>
#include <unordered_set>
#include <unordered_map>
#include <array>
#include <cassert>
#include <bitset>
using namespace std;
using LL = long long;

int N, M;
LL a[555], imos[555];
long double dp[555][555];

int main(void)
{
	cin >> N >> M;
	for (int i = 0; i < N; ++i) {
		cin >> a[i];
		imos[i + 1] = imos[i] + a[i];
	}
	for (int i = 0; i < 555; ++i) {
		for (int j = 0; j < 555; ++j) {
			dp[i][j] = -123456789012345678;
		}
	}
	dp[0][0] = 0;
	for (int i = 0; i < N; ++i) {
		for (int j = 0; j < M; ++j) {
			for (int k = i + 1; k <= N; ++k) {
				long double width = k - i;
				dp[k][j + 1] = max(dp[k][j + 1], dp[i][j] + (long double)(imos[k] - imos[i]) / width);
			}
		}
	}
	cout.precision(15);
	long double ans = 0.;
	for (int i = 1; i <= M; ++i) {
		ans = max(ans, dp[N][i]);
	}
	cout << ans << endl;
	return 0;
}

