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;
#define int long long
#define stoi stoll
using ll=long long;
using vi=vector<int>;
using pii=pair<int,int>;
#define ALL(c) begin(c),end(c)
#define RALL(c) rbegin(c),rend(c)
#define ITR(i,b,e) for(auto i=(b);i!=(e);++i)
#define FORE(x,c) for(auto &x:c)
#define REPF(i,a,n) for(int i=a,i##len=(int)(n);i<i##len;++i)
#define REP(i,n) REPF(i,0,n)
#define REPR(i,n) for(int i=(int)(n);i>=0;--i)
#define SZ(c) ((int)c.size())
#define CONTAIN(c,x) (c.find(x)!=end(c))
#define OUTOFRANGE(y,x,h,w) ((y)<0||(x)<0||(y)>=(h)||(x)>=(w))
#define dump(...)
const signed INF_=1001001001; const long long INF=1001001001001001001LL;
const int DX[9]={0,1,0,-1,1,1,-1,-1,0},DY[9]={-1,0,1,0,-1,1,1,-1,0};
template<class T> ostream& operator<<(ostream &os,const vector<T> &v) {
    ITR(i,begin(v),end(v))os<<*i<<(i==end(v)-1?"":" ");return os;}
template<class T> istream& operator>>(istream &is,vector<T> &v) {
    ITR(i,begin(v),end(v)) is>>*i;return is;}
template<class T,class U> istream& operator>>(istream &is, pair<T,U> &p) {
    is>>p.first>>p.second;return is;}
template<class T> bool chmax(T &a,const T &b){if(a<b){a=b;return 1;}return 0;}
template<class T> bool chmin(T &a,const T &b){if(a>b){a=b;return 1;}return 0;}
template<class T> using heap=priority_queue<T,vector<T>,greater<T>>;
struct{template<class T> operator T(){T x;cin>>x;return x;}} IN;
struct before_main_function {
    before_main_function() {
        // cin.tie(0);ios::sync_with_stdio(false);
        cout<<setprecision(15)<<fixed;
        // #define endl "\n"
    }
} before_main_function;
//------------------8<------------------------------------8<--------------------

double dp[600][600];
signed main() {
    REP(i, 600) REP(j, 600) dp[i][j] = -INF;

    int n, m;
    cin >> n >> m;
    vector<int> a(n);
    cin >> a;
    vector<int> sum(n+1, 0);
    REPF(i, 1, n+1) sum[i] = sum[i-1] + a[i-1];
    
    dp[0][0] = 0;
    REP(i, n+1) {
        REP(j, i+1) {
            REPF(k, 1, n+1) {
                if (i+k > n) continue;
                double avr = sum[i+k] - sum[i];
                avr /= k;
                chmax(dp[i+k][j+1], dp[i][j] + avr);
            }
        }
    }
    double ans = 0;
    REPF(i, 1, m+1) chmax(ans, dp[n][i]);
    cout << ans << endl;
    return 0;
}

