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 = 200005;
const long double eps = 1e-14;
double a[N];
int n;
int dcmp(long double a, long double b) {
  if (fabs(a - b) < eps) return 0;
  if (a > b)
    return 1;
  else
    return -1;
}
long double calc(long double x) {
  long double now = 0.0;
  long double ans1 = 0, ans2;
  for (int i = 1; i <= n; i++) {
    if (dcmp(now, 0.0) < 0) now = 0.0;
    now += a[i] - x;
    if (dcmp(now, ans1) > 0) ans1 = now;
  }
  now = 0.0;
  for (int i = 1; i <= n; i++) {
    if (dcmp(now, 0.0) > 0) now = 0.0;
    now += a[i] - x;
    if (dcmp(now, ans2) < 0) ans2 = now;
  }
  return max(abs(ans1), abs(ans2));
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lf", &a[i]);
  }
  long double l = -10000.0, r = 10000.0;
  long double l1, r1, al, ar;
  while (dcmp(l, r)) {
    l1 = (2 * l + r) / 3;
    r1 = (l + 2 * r) / 3;
    al = calc(l1);
    ar = calc(r1);
    if (dcmp(al, ar) >= 0) {
      l = l1;
    } else {
      r = r1;
    }
  }
  cout << setiosflags(ios::showpoint);
  cout << setprecision(15);
  cout << calc(l) << endl;
  cin >> n;
  return 0;
}