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>
double f_abs(double x) { return x < 0 ? -x : x; }
double f_max(double a, double b) { return a > b ? a : b; }
double f_min(double a, double b) { return a < b ? a : b; }
const double eps = 1e-12;
const int N = 200003;
int n;
double a[N], b[N];
void GetData() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%lf", &a[i]);
}
double Calc(double x) {
  int i;
  double dp, cur, res;
  for (i = 0; i < n; ++i) b[i] = a[i] - x;
  dp = cur = b[0];
  for (i = 1; i < n; ++i) {
    if (cur + b[i] < b[i])
      cur = b[i];
    else
      cur += b[i];
    dp = f_max(dp, cur);
  }
  res = f_abs(dp);
  dp = cur = b[0];
  for (i = 1; i < n; ++i) {
    if (cur + b[i] > b[i])
      cur = b[i];
    else
      cur += b[i];
    dp = f_min(dp, cur);
  }
  res = f_max(res, f_abs(dp));
  return res;
}
void Solve() {
  int cnt = 100;
  double lo, hi, midl, midr;
  lo = -1e4;
  hi = 1e4;
  while (fabs(hi - lo) > eps && cnt--) {
    midl = (hi + lo * 2) / 3;
    midr = (hi * 2 + lo) / 3;
    if (Calc(midl) > Calc(midr))
      lo = midl;
    else
      hi = midr;
  }
  printf("%.10lf\n", Calc(lo));
}
int main() {
  GetData();
  Solve();
  return 0;
}