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;
long double findWeakness(int arr[], int n, long double x) {
  long double minimum = 0, maximum = 0, ans = 0, sum = 0;
  for (int i = 0; i < n; i++) {
    sum += (arr[i] - x);
    long double currBest;
    if (sum < 0)
      currBest = sum - maximum;
    else
      currBest = sum - minimum;
    ans = max(ans, abs(currBest));
    if (sum < minimum) minimum = sum;
    if (sum > maximum) maximum = sum;
  }
  return ans;
}
long double weaknessAndPoorness(int arr[], int n) {
  long double start = INT_MIN, end = INT_MAX, precision = 1e-9, ans = 1e18;
  for (int i = 0; i < 300; i++) {
    long double midLeft = start + (end - start) / 3,
                midRight = end - (end - start) / 3;
    long double costLeft = findWeakness(arr, n, midLeft),
                costRight = findWeakness(arr, n, midRight);
    ans = min(ans, min(costLeft, costRight));
    if (costLeft < costRight)
      end = midRight;
    else
      start = midLeft;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  int arr[n];
  for (int i = 0; i < n; i++) cin >> arr[i];
  long double ans = weaknessAndPoorness(arr, n);
  printf("%.12Lf", ans);
  return 0;
}