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>
const double EPS = 1e-7, PI = acos(-1.0);
using namespace std;
template <class T>
inline T IN(T &num) {
  num = 0;
  char c = getchar(), f = 0;
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  if (c == '-') f = 1, c = getchar();
  while ('0' <= c && c <= '9') num = num * 10 + c - '0', c = getchar();
  if (f) num = -num;
  return num;
}
const int NUM = 200010;
int n;
double a[NUM];
double getRes(double x) {
  double res = 0, maxv = -1e16, minv = 1e16;
  for (int i = 0; i < n; ++i)
    maxv = max(maxv, res = max(a[i] - x, res + a[i] - x));
  res = 0;
  for (int i = 0; i < n; ++i)
    minv = min(minv, res = min(a[i] - x, res + a[i] - x));
  return max(abs(minv), abs(maxv));
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%lf", &a[i]);
  double l = -10000, r = 10000, ll, rr;
  while (true) {
    ll = (l + l + r) / 3.0;
    rr = (l + r + r) / 3.0;
    double ans1 = getRes(ll);
    double ans2 = getRes(rr);
    if (fabs(ans1 - ans2) < 1e-7 && fabs(l - r) < 1e-11) {
      printf("%.8f\n", ans1);
      return 0;
    }
    if (ans1 - ans2 < -EPS)
      r = rr;
    else
      l = ll;
  }
  return 0;
}