Problem: Radewoosh is playing a computer game. There are n levels, numbered 1 through n. Levels are divided into k regions (groups). Each region contains some positive number of consecutive levels.

The game repeats the the following process:

  1. If all regions are beaten then the game ends immediately. Otherwise, the system finds the first region with at least one non-beaten level. Let X denote this region.
  2. The system creates an empty bag for tokens. Each token will represent one level and there may be many tokens representing the same level.
    * For each already beaten level i in the region X, the system adds ti tokens to the bag (tokens representing the i-th level). 
    * Let j denote the first non-beaten level in the region X. The system adds tj tokens to the bag. 
  3. Finally, the system takes a uniformly random token from the bag and a player starts the level represented by the token. A player spends one hour and beats the level, even if he has already beaten it in the past. 



Given n, k and values t1, t2, ..., tn, your task is to split levels into regions. Each level must belong to exactly one region, and each region must contain non-empty consecutive set of levels. What is the minimum possible expected number of hours required to finish the game?

Input

The first line of the input contains two integers n and k (1 ≤ n ≤ 200 000, 1 ≤ k ≤ min(50, n)) — the number of levels and the number of regions, respectively.

The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 100 000).

Output

Print one real number — the minimum possible expected value of the number of hours spent to finish the game if levels are distributed between regions in the optimal way. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 4.

Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct if <image>.

Examples

Input

4 2
100 3 5 7


Output

5.7428571429


Input

6 2
1 2 4 8 16 32


Output

8.5000000000

Note

In the first sample, we are supposed to split 4 levels into 2 regions. It's optimal to create the first region with only one level (it must be the first level). Then, the second region must contain other three levels.

In the second sample, it's optimal to split levels into two regions with 3 levels each.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int INF = 2e9;
const long long LINF = 2e16;
const int magic = 348;
const int MOD = 1e9 + 7;
const double eps = 1e-12;
const double pi = acos(-1);
struct fastio {
  static const int S = 1e7;
  char rbuf[S + 48], wbuf[S + 48];
  int rpos, wpos, len;
  fastio() { rpos = len = wpos = 0; }
  inline char Getchar() {
    if (rpos == len) rpos = 0, len = fread(rbuf, 1, S, stdin);
    if (!len) return EOF;
    return rbuf[rpos++];
  }
  template <class T>
  inline void Get(T &first) {
    char ch;
    bool f;
    T res;
    while (!isdigit(ch = Getchar()) && ch != '-') {
    }
    if (ch == '-')
      f = false, res = 0;
    else
      f = true, res = ch - '0';
    while (isdigit(ch = Getchar())) res = res * 10 + ch - '0';
    first = (f ? res : -res);
  }
  inline void getstring(char *s) {
    char ch;
    while ((ch = Getchar()) <= 32) {
    }
    for (; ch > 32; ch = Getchar()) *s++ = ch;
    *s = '\0';
  }
  inline void flush() {
    fwrite(wbuf, 1, wpos, stdout);
    fflush(stdout);
    wpos = 0;
  }
  inline void Writechar(char ch) {
    if (wpos == S) flush();
    wbuf[wpos++] = ch;
  }
  template <class T>
  inline void Print(T first, char ch) {
    char s[20];
    int pt = 0;
    if (first == 0)
      s[++pt] = '0';
    else {
      if (first < 0) Writechar('-'), first = -first;
      while (first) s[++pt] = '0' + first % 10, first /= 10;
    }
    while (pt) Writechar(s[pt--]);
    Writechar(ch);
  }
  inline void printstring(char *s) {
    int pt = 1;
    while (s[pt] != '\0') Writechar(s[pt++]);
  }
} io;
template <typename T>
inline void check_max(T &first, T cmp) {
  first = max(first, cmp);
}
template <typename T>
inline void check_min(T &first, T cmp) {
  first = min(first, cmp);
}
template <typename T>
inline T myabs(T first) {
  return first >= 0 ? first : -first;
}
template <typename T>
inline T gcd(T first, T second) {
  return second == 0 ? first : gcd(second, first % second);
}
inline int add(int first) {
  if (first >= MOD) first -= MOD;
  return first;
}
inline int add(int first, int MO) {
  if (first >= MO) first -= MO;
  return first;
}
inline int sub(int first) {
  if (first < 0) first += MOD;
  return first;
}
inline int sub(int first, int MO) {
  if (first < 0) first += MO;
  return first;
}
inline void Add(int &first, int second) { first = add(first + second); }
inline void Add(int &first, int second, int MO) {
  first = add(first + second, MO);
}
inline void Sub(int &first, int second) { first = sub(first - second); }
inline void Sub(int &first, int second, int MO) {
  first = sub(first - second, MO);
}
template <typename T>
inline int quick_pow(int first, T second) {
  int res = 1;
  while (second) {
    if (second & 1) res = 1ll * res * first % MOD;
    first = 1ll * first * first % MOD;
    second >>= 1;
  }
  return res;
}
template <typename T>
inline int quick_pow(int first, T second, int MO) {
  int res = 1;
  while (second) {
    if (second & 1) res = 1ll * res * first % MO;
    first = 1ll * first * first % MO;
    second >>= 1;
  }
  return res;
}
const int MAXN = 2e5;
inline bool islarger(double first, double second) {
  return first - second > eps;
}
inline bool issmaller(double first, double second) {
  return first - second < -eps;
}
inline bool isequal(double first, double second) {
  return (!islarger(first, second)) && (!issmaller(first, second));
}
int n, k;
int t[MAXN + 48];
double dp[51][MAXN + 48];
double p[MAXN + 48], q[MAXN + 48], r[MAXN + 48];
inline void init() {
  for (register int i = 1; i <= n; i++) p[i] = p[i - 1] + t[i];
  for (register int i = 1; i <= n; i++) q[i] = q[i - 1] + double(1.0) / t[i];
  for (register int i = 1; i <= n; i++)
    r[i] = r[i - 1] + double(1.0) * double(t[i]) * q[i - 1];
}
inline double calc_k(double x1, double y1, double x2, double y2) {
  if (isequal(x1, x2)) return 1e16;
  return (y1 - y2) / (x1 - x2);
}
inline double calc(int left, int right) {
  return p[right] * q[right] - p[left - 1] * q[right] - r[right] + r[left - 1];
}
struct node {
  double first, second;
  int ind;
  node() {}
  inline node(double _x, double _y, int _i) {
    first = _x;
    second = _y;
    ind = _i;
  }
} Q[MAXN + 48];
int head, tail;
int main() {
  scanf("%d%d", &n, &k);
  for (register int i = 1; i <= n; i++) scanf("%d", t + i);
  init();
  for (register int i = 1; i <= n; i++) dp[1][i] = calc(1, i);
  for (register int layer = 2; layer <= k; layer++) {
    head = tail = 1;
    Q[1] =
        node(p[layer - 1], dp[layer - 1][layer - 1] + r[layer - 1], layer - 1);
    for (register int i = layer; i <= n; i++) {
      while (head < tail &&
             issmaller(calc_k(Q[head].first, Q[head].second, Q[head + 1].first,
                              Q[head + 1].second),
                       q[i]))
        head++;
      int chosen = Q[head].ind;
      dp[layer][i] = dp[layer - 1][chosen] + calc(chosen + 1, i);
      Q[++tail] = node(p[i], dp[layer - 1][i] + r[i], i);
      while (head + 1 < tail &&
             issmaller(calc_k(Q[tail - 1].first, Q[tail - 1].second,
                              Q[tail].first, Q[tail].second),
                       calc_k(Q[tail - 2].first, Q[tail - 2].second,
                              Q[tail - 1].first, Q[tail - 1].second)))
        Q[tail - 1] = Q[tail], tail--;
    }
  }
  printf("%.10lf\n", dp[k][n]);
  io.flush();
  return 0;
}