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;
template <class T>
inline bool getmin(T *a, const T &b) {
  if (b < *a) {
    *a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool getmax(T *a, const T &b) {
  if (b > *a) {
    *a = b;
    return true;
  }
  return false;
}
template <class T>
inline void read(T *a) {
  char c;
  while (isspace(c = getchar())) {
  }
  bool flag = 0;
  if (c == '-')
    flag = 1, *a = 0;
  else
    *a = c - 48;
  while (isdigit(c = getchar())) *a = *a * 10 + c - 48;
  if (flag) *a = -*a;
}
const int mo = 1000000007;
template <class T>
T pow(T a, T b, int c = mo) {
  T res = 1;
  for (T i = 1; i <= b; i <<= 1, a = 1LL * a * a % c)
    if (b & i) res = 1LL * res * a % c;
  return res;
}
const int N = 210000;
double a[N];
int n;
double check(double x) {
  double tmp = 0, res = 0;
  for (int i = (1); i <= (n); ++i) {
    tmp = max(a[i] - x, tmp + a[i] - x);
    getmax(&res, tmp);
  }
  tmp = 0;
  for (int i = (1); i <= (n); ++i) {
    tmp = min(a[i] - x, tmp + a[i] - x);
    getmax(&res, -tmp);
  }
  return res;
}
int main() {
  cin >> n;
  for (int i = (1); i <= (n); ++i) read(a + i);
  double l = -2e9, r = 2e9, ans = 2e9;
  for (int tt = (1); tt <= (200); ++tt) {
    double m1 = (l + r) / 2;
    double m2 = (m1 + r) / 2;
    double fl = check(l), fr = check(r), f1 = check(m1), f2 = check(m2);
    getmin(&ans, fl);
    getmin(&ans, f1);
    getmin(&ans, f2);
    getmin(&ans, fr);
    if (f1 < f2)
      r = m2;
    else
      l = m1;
  }
  printf("%.10lf\n", ans);
  return 0;
}