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 = 100 * 1000 * 2 + 10;
int arr[N], n;
double solver(double d, bool reverse) {
  double ret = 0;
  double curr = 0;
  for (int i = 0; i < n; i++) {
    curr += (arr[i] - d) * pow(-1, reverse);
    if (curr < 0) curr = 0;
    ret = max(ret, curr);
  }
  return ret;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i]);
  }
  double ini = -200 * 100 - 2;
  double fim = 200 * 100 + 2;
  int interacoes = 200;
  while (interacoes--) {
    double terco = (fim - ini) / 3.0;
    double ter1 = ini + terco;
    double ter2 = fim - terco;
    double resp1 = max(solver(ter1, false), solver(ter1, true));
    double resp2 = max(solver(ter2, false), solver(ter2, true));
    if (resp1 < resp2) {
      fim = ter2;
    } else {
      ini = ter1;
    }
  }
  printf("%.10f", max(solver((ini + fim) / 2.0, false),
                      solver((ini + fim) / 2.0, true)));
  return 0;
}