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 INF = 1000000000;
const long long INFll = 1000000000000000000ll;
const int MOD = 1000000007;
const long double EPS = 0.0000000000001;
int n, m;
int a[200500];
long double s;
inline long double getMin(long double x) {
  long double ret = 0, t = 0;
  long double mn, mx;
  s = 0;
  mn = 0;
  mx = -INFll;
  for (int i = 0; i < n; ++i) {
    s += a[i] - x;
    mx = max(mx, s - mn);
    mn = min(s, mn);
  }
  mx = abs(mx);
  return mx;
}
inline long double getMax(long double x) {
  long double ret = 0, t = 0;
  long double mn, mx;
  s = 0;
  mn = INFll;
  mx = 0;
  for (int i = 0; i < n; ++i) {
    s += a[i] - x;
    mn = min(mn, s - mx);
    mx = max(mx, s);
  }
  mn = abs(mn);
  return mn;
}
inline long double get(long double x) { return max(getMin(x), getMax(x)); }
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  long double l = -10000;
  long double r = 10000;
  while (r - l > 5 * EPS) {
    long double m1 = l + (r - l) / 3.;
    long double m2 = r - (r - l) / 3.;
    long double f1 = get(m1);
    long double f2 = get(m2);
    if (f1 > f2) {
      l = m1;
    } else {
      r = m2;
    }
  }
  long double ans = INFll;
  while (l <= r) {
    ans = min(ans, get(l));
    l += EPS;
  }
  cout << fixed << setprecision(6) << ans << endl;
  return 0;
}