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];
int n;
double cal(double x) {
  double t1 = 0, t2 = 0, maxsum = -1320, minsum = 1320;
  for (int i = 1; i <= n; i++) {
    if (t1 >= 0) {
      t1 += a[i] - x;
    } else {
      t1 = a[i] - x;
    }
    if (t2 <= 0) {
      t2 += a[i] - x;
    } else {
      t2 = a[i] - x;
    }
    maxsum = max(maxsum, t1);
    minsum = min(minsum, t2);
  }
  return max(fabs(maxsum), fabs(minsum));
}
double triple(double L, double R) {
  int t = 100;
  while (t--) {
    double mid = L + (R - L) / 3.0;
    double mmid = R - (R - L) / 3.0;
    if (cal(mid) > cal(mmid))
      L = mid;
    else
      R = mmid;
  }
  return cal(L) > cal(R) ? L : R;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  printf("%.7lf\n", cal(triple(-10005, 10005)));
}