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;
using ll = long long;
const ll llinf = 0x3f3f3f3f3f3f3f3f;
const ll inf = 0x3f3f3f3f3f;
const double eps = 1e-9;
int n;
int t[200010];
double now, max_glo, min_glo;
double a, b, m1, m2, res;
double f(double x) {
  now = 0;
  max_glo = -llinf;
  for (int i = 0; i < n; i++) {
    if (0 > now) {
      now = t[i] - x;
    } else {
      now += t[i] - x;
    }
    if (max_glo < now) {
      max_glo = now;
    }
  }
  now = 0;
  min_glo = llinf;
  for (int i = 0; i < n; i++) {
    if (0 < now) {
      now = t[i] - x;
    } else {
      now += t[i] - x;
    }
    if (min_glo > now) {
      min_glo = now;
    }
  }
  return max(abs(max_glo), abs(min_glo));
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> t[i];
  int mx = -inf;
  int mn = inf;
  for (int i = 0; i < n; i++) {
    mx = max(mx, t[i]);
    mn = min(mn, t[i]);
  }
  a = mn;
  b = mx;
  int cnt = 0;
  while (cnt++ < 200) {
    m1 = a + (b - a) / 3;
    m2 = b - (b - a) / 3;
    if (f(m1) < f(m2)) {
      b = m2;
    } else {
      a = m1;
    }
  }
  cout << fixed << f(b) << '\n';
  return 0;
}