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 a[200000], n;
pair<double, double> go(double x) {
  double sum1 = 0, sum2 = 0, mx1 = 0, mx2 = 0;
  for (int i = 0; i < (int)(n); i++) {
    sum1 += a[i] - x;
    if (sum1 < 0) sum1 = 0;
    mx1 = max(mx1, sum1);
    sum2 += a[i] - x;
    if (sum2 > 0) sum2 = 0;
    mx2 = min(mx2, sum2);
  }
  return make_pair(mx1, mx2);
}
int main() {
  int mx = INT_MIN, mn = INT_MAX;
  cin >> n;
  for (int i = 0; i < (int)(n); i++) {
    scanf("%d", &a[i]);
    a[i] += 10000;
    mx = max(mx, a[i]);
    mn = min(mn, a[i]);
  }
  double l = mn, r = mx, ans = l;
  for (int i = 0; i < (int)(100); i++) {
    double m = (r + l) / 2.0;
    pair<double, double> qq = go(m);
    if (fabs(fabs(qq.first) - fabs(qq.second)) < 1e-9) {
      ans = m;
      break;
    } else if (fabs(qq.first) > fabs(qq.second))
      l = m, ans = m;
    else
      r = m, ans = m;
  }
  printf("%.9lf\n", fabs(go(ans).first));
}