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;
static const int MAXN = 2e5 + 10;
static const int EPS = 1e-8;
int n;
int data[MAXN];
double arr[MAXN];
double Cal() {
  double tp = 0;
  double ret = arr[1];
  for (int i = 1; i <= n; ++i) {
    tp += arr[i];
    ret = max(tp, ret);
    if (tp < EPS + 0.0) {
      tp = 0;
    }
  }
  return ret;
}
double Solve(double x) {
  for (int i = 1; i <= n; ++i) arr[i] = data[i] - x;
  double tp1 = Cal();
  for (int i = 1; i <= n; ++i) arr[i] = -arr[i];
  double tp2 = Cal();
  double ret = max(fabs(tp1), fabs(tp2));
  return ret;
}
int main() {
  scanf("%d", &n);
  int mi = 10001, mx = -10001;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &data[i]);
    mi = min(data[i], mi);
    mx = max(data[i], mx);
  }
  double l = mi, r = mx;
  for (int i = 1; i <= 100; ++i) {
    double ll = (2 * l + r) / 3.0, rr = (2 * r + l) / 3.0;
    double tp1 = Solve(ll), tp2 = Solve(rr);
    if (tp1 > tp2) {
      l = ll;
    } else
      r = rr;
  }
  printf("%.7f", Solve(l));
}