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 double eps = 1e-8;
const int INF = (1 << 30) - 1;
const int MAXN = 200010;
int n;
int A[MAXN];
double B[MAXN];
double sum[MAXN];
double Cal(double v) {
  for (int i = 1; i <= n; ++i) {
    B[i] = 1.0 * A[i] - v;
  }
  long long MAX_LL = 1LL << 60;
  double sum_max = 0, sum_min = 0;
  double ans_max = 0, ans_min = MAX_LL;
  for (int i = 1; i <= n; ++i) {
    if (sum_min > 0) sum_min = 0;
    if (sum_max < 0) sum_max = 0;
    sum_max += B[i];
    sum_min += B[i];
    ans_max = max(ans_max, sum_max);
    ans_min = min(ans_min, sum_min);
  }
  if (ans_max < 0) ans_max = -ans_max;
  if (ans_min < 0) ans_min = -ans_min;
  return max(ans_max, ans_min);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &A[i]);
  }
  double l = -10000.0, r = 10000.0;
  int cnt = 100;
  while (cnt--) {
    double mid = ((l) + ((r) - (l)) / 2);
    double MID = ((mid) + ((r) - (mid)) / 2);
    double A1 = Cal(mid);
    double A2 = Cal(MID);
    if (A1 < A2)
      r = MID;
    else
      l = mid;
  }
  printf("%.12f\n", Cal(l));
  return 0;
}