Problem: You are given a sequence of n integers a1, a2, ..., an. 

Determine a real number x such that the weakness of the sequence a1 - x, a2 - x, ..., an - x is as small as possible.

The weakness of a sequence is defined as the maximum value of the poorness over all segments (contiguous subsequences) of a sequence.

The poorness of a segment is defined as the absolute value of sum of the elements of segment.

Input

The first line contains one integer n (1 ≤ n ≤ 200 000), the length of a sequence.

The second line contains n integers a1, a2, ..., an (|ai| ≤ 10 000).

Output

Output a real number denoting the minimum possible weakness of a1 - x, a2 - x, ..., an - x. Your answer will be considered correct if its relative or absolute error doesn't exceed 10 - 6.

Examples

Input

3
1 2 3


Output

1.000000000000000


Input

4
1 2 3 4


Output

2.000000000000000


Input

10
1 10 2 9 3 8 4 7 5 6


Output

4.500000000000000

Note

For the first case, the optimal value of x is 2 so the sequence becomes  - 1, 0, 1 and the max poorness occurs at the segment "-1" or segment "1". The poorness value (answer) equals to 1 in this case. 

For the second sample the optimal value of x is 2.5 so the sequence becomes  - 1.5, - 0.5, 0.5, 1.5 and the max poorness occurs on segment "-1.5 -0.5" or "0.5 1.5". The poorness value (answer) equals to 2 in this case.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int dr[] = {1, 0, -1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, -1, 1};
const double eps = 1e-9;
const int INF = 0x7FFFFFFF;
const long long INFLL = 0x7FFFFFFFFFFFFFFFLL;
const double pi = acos(-1);
template <class T>
T take(queue<T> &O) {
  T tmp = O.front();
  O.pop();
  return tmp;
}
template <class T>
T take(stack<T> &O) {
  T tmp = O.top();
  O.pop();
  return tmp;
}
template <class T>
T take(priority_queue<T> &O) {
  T tmp = O.top();
  O.pop();
  return tmp;
}
template <class T>
inline void getint(T &num) {
  bool neg = 0;
  num = 0;
  char c;
  while ((c = getchar_unlocked()) && (!isdigit(c) && c != '-'))
    ;
  if (c == '-') {
    neg = 1;
    c = getchar_unlocked();
  }
  do num = num * 10 + c - '0';
  while ((c = getchar_unlocked()) && isdigit(c));
  if (neg) num = -num;
}
template <class T>
inline bool inRange(T z, T a, T b) {
  return a <= z && z <= b;
}
void OPEN(string in = "input.txt", string out = "output.txt") {
  freopen(in.c_str(), "r", stdin);
  freopen(out.c_str(), "w", stdout);
  return;
}
int arr[200005];
int n;
vector<int> vc;
inline double cek(double x) {
  double tmp = 0.0, ansz = 0.0;
  for (int(i) = (0), _t = (n); i < (_t); ++(i)) {
    tmp = max(tmp + arr[i] - x, 0.0);
    ansz = max(ansz, tmp);
  }
  tmp = 0.0;
  for (int(i) = (0), _t = (n); i < (_t); ++(i)) {
    tmp = max(tmp + -(arr[i] - x), 0.0);
    ansz = max(ansz, tmp);
  }
  return ansz;
}
int main(int argc, const char *argv[]) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int(i) = (0), _t = (n); i < (_t); ++(i)) cin >> arr[i];
  double lo = -10000.0, hi = 10000.0, le, ri;
  for (int(i) = (0), _t = (120); i < (_t); ++(i)) {
    le = lo + (hi - lo) / 3.;
    ri = lo + (hi - lo) * 2. / 3.;
    double cle = cek(le), cri = cek(ri);
    if (cle < cri)
      hi = ri;
    else if (cle > cri)
      lo = le;
    else {
      lo = le;
      hi = ri;
    }
  }
  cout << fixed << setprecision(10) << cek(lo) << "\n";
  return 0;
}