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>
long long mpow(long long a, long long n, long long mod) {
  long long ret = 1;
  long long b = a;
  while (n) {
    if (n & 1) ret = (ret * b) % mod;
    b = (b * b) % mod;
    n >>= 1;
  }
  return (long long)ret;
}
using namespace std;
int a[2 * 100000 + 5];
int n;
int i, j;
double X(double x) {
  vector<double> v;
  for (int j = 0; j < n; j++) v.push_back(a[j] - x);
  double curmax = 0;
  double sum = 0;
  for (j = 0; j < n; j++) {
    sum += v[j];
    if (curmax < 0 && sum >= 0) {
      if (abs(curmax) < abs(sum - curmax)) {
        double k = curmax;
        curmax = sum - k;
        sum = sum - k;
      }
    } else if (curmax >= 0 && sum <= 0) {
      if (abs(curmax) < abs(-curmax + sum)) {
        double k = curmax;
        curmax = -k + sum;
        sum = -k + sum;
      }
    } else {
      if (abs(curmax) < abs(sum)) {
        curmax = sum;
      }
    }
  }
  return abs(curmax);
}
int main() {
  scanf("%d", &n);
  for (j = 0; j < n; j++) scanf("%d", &a[j]);
  double lo = -10001;
  double hi = 10001;
  i = 0;
  while (i != 300) {
    double g = lo + (hi - lo) / 3.0;
    double h = hi - (hi - lo) / 3.0;
    if (X(g) < X(h)) {
      hi = h;
    } else
      lo = g;
    i++;
  }
  cout << fixed << setprecision(10) << X(lo);
  return 0;
}