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;
const int N = 2e5 + 100;
const int inf = 1e9;
int n, a[N];
double ma(double p) {
  double sum = a[1] - p, b = a[1] - p;
  for (int i = 2; i <= n; i++) {
    if (b > 0)
      b += a[i] - p;
    else
      b = a[i] - p;
    sum = max(sum, b);
  }
  return abs(sum);
}
double mi(double p) {
  double sum = a[1] - p, b = a[1] - p;
  for (int i = 2; i <= n; i++) {
    if (b < 0)
      b += a[i] - p;
    else
      b = a[i] - p;
    sum = min(sum, b);
  }
  return abs(sum);
}
double work(double p) { return max(ma(p), mi(p)); }
double cal() {
  double l = -10000, r = -l;
  double X;
  int cnt = 300;
  while (cnt--) {
    double mid1 = (2 * l + r) / 3.0, mid2 = (2 * r + l) / 3;
    double x = work(mid1), y = work(mid2);
    X = min(x, y);
    if (x <= y) {
      r = mid2;
    } else {
      l = mid1;
    }
  }
  return X;
}
int main() {
  while (~scanf("%d", &n)) {
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    printf("%.12f\n", cal());
  }
  return 0;
}