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;
double a[200007];
int n;
double cal(double x) {
  double s, maxx, minn;
  maxx = a[0] - x;
  s = 0;
  for (int i = 0; i < n; i++) {
    if (s < 0)
      s = a[i] - x;
    else
      s = s + a[i] - x;
    if (s > maxx) maxx = s;
  }
  minn = a[0] - x;
  s = 0;
  for (int i = 0; i < n; i++) {
    if (s > 0)
      s = a[i] - x;
    else
      s = s + a[i] - x;
    if (s < minn) minn = s;
  }
  return max(fabs(maxx), fabs(minn));
}
int main() {
  while (scanf("%d", &n) == 1) {
    double L = 100000, R = -100000, mid, mmid, t1, t2, ans;
    for (int i = 0; i < n; i++) {
      scanf("%lf", a + i);
      L = min(a[i], L);
      R = max(a[i], R);
    }
    for (int i = 0; i < 200; i++) {
      mid = L + (R - L) / 3;
      mmid = R - (R - L) / 3;
      t1 = cal(mid);
      t2 = cal(mmid);
      if (t1 < t2)
        R = mmid, ans = t1;
      else
        L = mid, ans = t2;
    }
    printf("%.16f\n", ans);
  }
  return 0;
}