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;
int n, arr[200002];
void Input() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &arr[i]);
  }
}
long double eabs(long double num) {
  if (num < 0) return -num;
  return num;
}
long double Evaluate(long double x) {
  long double res = 0.0, curmin = 0.0, curmax = 0.0, cursum = 0.0;
  for (int i = 1; i <= n; i++) {
    long double val = (long double)arr[i] - x;
    cursum += val;
    curmin = min(curmin, cursum);
    curmax = max(curmax, cursum);
    res = max(res, eabs(cursum - curmin));
    res = max(res, eabs(cursum - curmax));
  }
  return res;
}
long double eps = 0.0000000000001;
bool Equal(long double a, long double b) { return (eabs(a - b) < eps); }
int main() {
  long double bl, br, bmid, bres = -1.0;
  bl = -10000.0;
  br = 10000.0;
  Input();
  while (bl < br || (Equal(bl, br))) {
    bmid = (bl + br) / 2.0;
    long double prev = Evaluate(bmid - eps), cur = Evaluate(bmid);
    if (bres < 0) bres = cur;
    bres = min(bres, cur);
    if (prev < cur) {
      br = bmid - eps;
    } else {
      bl = bmid + eps;
    }
  }
  printf("%.7f\n", (double)bres);
}