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 long long inf = (1ll << 50);
const int mod = inf + 7;
const double pi = acos(-1.0);
const double eps = 1e-10;
double mi[220000], ma[220000], u[220000];
double second[220000];
int n, a[220000];
double first(double x) {
  for (int i = 1; i <= n; i++) second[i] = second[i - 1] + a[i] - x;
  mi[n] = second[n];
  ma[n] = second[n];
  for (int i = n - 1; i >= 1; i--) {
    ma[i] = max(ma[i + 1], second[i]);
    mi[i] = min(mi[i + 1], second[i]);
  }
  double res = 0, su = 0;
  for (int i = 1; i <= n; i++) {
    res = max(ma[i] + su, res);
    res = max(-(mi[i] + su), res);
    su += -a[i] + x;
  }
  return res;
}
int main() {
  scanf("%d", &n);
  long long e = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    e += a[i];
  }
  double l = -(1e14);
  double r = (1e14);
  for (int i = 0; i <= 300; i++) {
    double m1 = l + (r - l) / 3.0;
    double m2 = r - (r - l) / 3.0;
    if (first(m1) < first(m2))
      r = m2;
    else
      l = m1;
  }
  printf("%.10lf", first(l));
  return 0;
}