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;
inline void Rd(int &res) {
  char c;
  res = 0;
  while (c = getchar(), c < '0')
    ;
  do {
    res = (res << 1) + (res << 3) + (c ^ 48);
  } while (c = getchar(), c >= '0');
}
const int M = 200005;
int A[M], n;
double sum[M];
double check(double x) {
  double res = 0;
  sum[0] = 0;
  for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + A[i] - x;
  double mx = 0, mi = 0;
  for (int i = 1; i <= n; i++) {
    if (mx - sum[i] > res) res = mx - sum[i];
    if (sum[i] - mi > res) res = sum[i] - mi;
    if (sum[i] > mx) mx = sum[i];
    if (sum[i] < mi) mi = sum[i];
  }
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &A[i]);
  double L = -20000, R = 20000, ans;
  for (int i = 1; i <= 300; i++) {
    double mid1 = L + (R - L) / 3;
    double mid2 = L + 2 * (R - L) / 3;
    double rs1 = check(mid1);
    double rs2 = check(mid2);
    if (rs1 < rs2) {
      ans = rs1;
      R = mid2;
    } else {
      ans = rs2;
      L = mid1;
    }
  }
  printf("%.15f\n", ans);
  return 0;
}