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[200005];
long double b[200005];
int n;
long double solve() {
  long double ans = 0;
  long double sum1 = 0, sum2 = 0;
  for (int i = 1; i <= n; i++) {
    sum1 = max(b[i], sum1 + b[i]);
    ans = max(ans, sum1);
    b[i] = -b[i];
    sum2 = max(b[i], sum2 + b[i]);
    ans = max(ans, sum2);
  }
  return ans;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  long double lo = -1e9, hi = 1e9, ans = 1e9;
  for (int i = 0; i < 200; i++) {
    long double mid1 = lo + (hi - lo) / 3;
    long double mid2 = hi - (hi - lo) / 3;
    long double res1 = -1e18, res2 = -1e18;
    for (int i = 1; i <= n; i++) {
      b[i] = a[i] - mid1;
    }
    res1 = solve();
    for (int i = 1; i <= n; i++) {
      b[i] = a[i] - mid2;
    }
    res2 = solve();
    if (res1 <= res2) {
      ans = min(ans, res1);
      hi = mid2;
    } else {
      ans = min(ans, res2);
      lo = mid1;
    }
  }
  printf("%.20Lf\n", ans);
}