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 b[300001], sum[300001];
int a[300001];
int n;
double check(double x) {
  for (int i = 1; i <= n; i++) b[i] = a[i] * 1.0 - x;
  for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + b[i];
  double Min = 0.0, Max = 0.0;
  double ans = 0.0;
  for (int i = 1; i <= n; i++) {
    ans = max(ans, fabs(sum[i] - Min));
    ans = max(ans, fabs(sum[i] - Max));
    Min = min(Min, sum[i]);
    Max = max(Max, sum[i]);
  }
  return ans;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  double l = -999999.0, r = 999999.0;
  double ans = 9999999999999.0;
  bool flag = false;
  for (int i = 1; i <= 200; i++) {
    double mid1 = l + (r - l) / 3.0, mid2 = r - (r - l) / 3.0;
    double tmpx = check(mid1), tmpy = check(mid2);
    if (tmpx > tmpy)
      l = mid1;
    else
      r = mid2;
    if (flag == false) {
      flag = true;
      ans = tmpx;
    } else
      ans = min(ans, tmpx);
    ans = min(ans, tmpy);
  }
  ans = min(ans, check(l));
  printf("%.9f\n", ans);
  return 0;
}