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, a[200000 + 3];
double prefix[200000 + 3];
pair<double, double> getWeakness(double x) {
  prefix[0] = 0;
  for (int i = 1; i <= N; i++) prefix[i] = prefix[i - 1] + (a[i] - x);
  double minPrefix = 0, maxPrefix = 0;
  double maxF = -1e10, maxG = -1e10;
  for (int j = 1; j <= N; j++) {
    double F = prefix[j] - minPrefix;
    double G = -prefix[j] + maxPrefix;
    maxF = max(maxF, F);
    maxG = max(maxG, G);
    minPrefix = min(minPrefix, prefix[j]);
    maxPrefix = max(maxPrefix, prefix[j]);
  }
  return {maxF, maxG};
}
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) scanf("%d", &a[i]);
  double lo = -10000, hi = 10000, mid;
  for (int rep = 1; rep <= 100; rep++) {
    mid = (lo + hi) / 2.0;
    auto weakness = getWeakness(mid);
    if (weakness.first > weakness.second)
      lo = mid;
    else
      hi = mid;
  }
  auto weakness = getWeakness(lo);
  printf("%.10lf\n", max(weakness.first, weakness.second));
}