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, m, o, re = 0;
double res = 0;
double val = 10000;
int a[1000010], b[1000010];
bool chk[1000010];
double calc(double d) {
  double ss = 0, sm = 0, ss2 = 0;
  for (int j = 0; j < n; j++) {
    if (ss < 0) ss = 0;
    if (ss2 > 0) ss2 = 0;
    ss += (double)a[j] - d;
    ss2 += (double)a[j] - d;
    sm = max(sm, max(((ss) < 0 ? -(ss) : (ss)), ((ss2) < 0 ? -(ss2) : (ss2))));
  }
  return sm;
}
int main() {
  int i, j, k, l;
  double ll = 0, rr = 20000, mid;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  for (i = 0; i < 100; i++) {
    mid = (ll + rr) / 2;
    res = calc(mid - val);
    double res2 = calc(mid - val + 1e-12);
    if (res < res2)
      rr = mid;
    else
      ll = mid;
  }
  printf("%.7lf", calc(ll - val));
  return 0;
}