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;
void RI() {}
template <typename... T>
void RI(int& head, T&... tail) {
  scanf("%d", &head);
  RI(tail...);
}
const int N = 200005, INF = 10005;
int num[N], n;
double findMaxSeq(double x, int t) {
  double Max = -INF, now = 0;
  for (int i = 0; i < n; i++) {
    now += t * (num[i] + x);
    Max = max(Max, now);
    if (now < 0) now = 0;
  }
  return Max;
}
double ABS(double f) { return f < 0 ? -f : f; }
int main() {
  RI(n);
  int flag = 1;
  for (int i = 0; i < n; i++) {
    RI(num[i]);
    if (i > 0 && num[i] != num[i - 1]) flag = 0;
  }
  if (n == 1 || flag == 1) {
    puts("0");
    return 0;
  }
  double L = -INF, R = INF, M, eps = 1e-9;
  int step = 100;
  while (R - L > eps || step != 0) {
    step--;
    M = (L + R) / 2;
    double Max = ABS(findMaxSeq(M, 1)), Min = ABS(findMaxSeq(M, -1));
    if (Max > Min)
      R = M;
    else
      L = M;
  }
  printf("%.8f\n", findMaxSeq((L + R) / 2, 1));
  return 0;
}