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 = 200000, MAX = 10000;
double A[N];
int n;
double calc() {
  double ret = 0, cur = 0;
  for (int i = 0; i < n; i++) {
    cur = max(cur + A[i], 0.0);
    ret = max(ret, cur);
  }
  return ret;
}
double f(double delta) {
  for (int i = 0; i < n; i++) A[i] += delta;
  double ret = calc();
  for (int i = 0; i < n; i++) A[i] *= -1;
  ret = max(ret, calc());
  for (int i = 0; i < n; i++) A[i] *= -1;
  for (int i = 0; i < n; i++) A[i] -= delta;
  return ret;
}
double search(double from, double to) {
  for (int i = 0; i < 100; i++) {
    double m1 = (2 * from + to) / 3, m2 = (from + 2 * to) / 3;
    if (f(m1) < f(m2))
      to = m2;
    else
      from = m1;
  }
  return from;
}
bool solve(int testNumber) {
  if (scanf("%d", &n) == EOF) return false;
  for (int i = 0; i < n; i++) scanf("%lf", &A[i]);
  printf("%.12f\n", f(search(-MAX, MAX)));
  return true;
}
int main() {
  for (int i = 1; solve(i); i++)
    ;
  return 0;
}