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 ll = long long;
using P = pair<int,int>;
#define enld '\n'
#define rep(i,n) for(int i=0; i<(n); i++)
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("Ofast")
constexpr ll INF = 1e18;
constexpr int inf = 1e9;
constexpr ll mod = 1000000007;
constexpr ll mod2 = 998244353;
const double PI = 3.1415926535897932384626433832795028841971;
const int dx[6] = {1, 0, -1, 0,1,1};
const int dy[6] = {0, 1, 0, -1,1,-1};
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
// ---------------------------------------------------------------------------


int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int N,M;
    cin >> N >> M;
    vector<ll> A(N);
    for(int i=0; i<N; i++){
        cin >> A[i];
    }
    vector<vector<double>> dp(N+1, vector<double> (M+2,0));
    double s = 0;
    for(int i=0; i<N; i++){
        s += A[i];
        dp[i+1][1] = s/(i+1);
    }
    for(int i=1; i<N; i++){
        for(int j=1; j<M; j++){
            if(dp[i][j] == 0) break;
            double sum = 0;
            for(int k=i+1; k<=N; k++){
                sum += A[k-1];
                chmax(dp[k][j+1],dp[i][j]+sum/(k-i));
            }
        }
    }
    // for(int i=0; i<=N; i++){
    //     for(int j=0; j<=M; j++){
    //         cout << dp[i][j] << " \n"[j==M];
    //     }
    // }
    cout << fixed << setprecision(10) << dp[N][M] << "\n";
    return 0;
}

