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 = 2 * 100 * 1000;
long double l[N];
long double Maks(long double dod, int n) {
  int i;
  long double as, mi, w;
  mi = 0;
  w = -10000000000000;
  as = 0;
  for (i = 1; i <= n; ++i) {
    as += (l[i] - dod);
    w = max(w, as - mi);
    mi = min(mi, as);
  }
  return w;
}
long double Minim(long double dod, int n) {
  int i;
  long double as, ma, w;
  ma = 0;
  w = 10000000000000;
  as = 0;
  for (i = 1; i <= n; ++i) {
    as += (l[i] - dod);
    w = min(w, as - ma);
    ma = max(ma, as);
  }
  return w;
}
long double TS(long double p, long double k, int n) {
  int i;
  long double s1, s2, w1, w2;
  for (i = 1; i <= 100; ++i) {
    s1 = p + ((k - p) / 3);
    s2 = k - ((k - p) / 3);
    w1 = max(abs(Maks(s1, n)), abs(Minim(s1, n)));
    w2 = max(abs(Maks(s2, n)), abs(Minim(s2, n)));
    if (w1 < w2)
      k = s2;
    else
      p = s1;
  }
  return p;
}
void WczytajLiczby(int &n) {
  int i;
  cin >> n;
  for (i = 1; i <= n; ++i) cin >> l[i];
}
void Weakness() {
  int n;
  long double w;
  WczytajLiczby(n);
  w = TS(-10000, 10000, n);
  w = max(abs(Minim(w, n)), abs(Maks(w, n)));
  cout << fixed << setprecision(10) << w << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  Weakness();
  return 0;
}