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>
const long long mod = 1000000007ll;
using namespace std;
int ar[200007];
int n;
double maxe, mine, mxtmp, mntmp;
inline void f(double x) {
  mine = 10000;
  maxe = -10000;
  mxtmp = mntmp = 0;
  for (int i = 0; i < n; i++) {
    mxtmp += (ar[i] - x);
    if (mxtmp < 0) mxtmp = 0;
    if (mxtmp > maxe) maxe = mxtmp;
    mntmp += (ar[i] - x);
    if (mntmp > 0) mntmp = 0;
    if (mntmp < mine) mine = mntmp;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", ar + i);
  }
  double l = -10000, h = 10000, m;
  double minweak = 10000000000;
  int lim = 100;
  while (lim--) {
    m = (l + h) / 2;
    f(m);
    if (maxe + mine > 0)
      l = m;
    else
      h = m;
  }
  printf("%.9f\n", maxe);
  return 0;
}