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;
long double eps = 1e-13;
int data[200010];
long double dp[200010][2];
int n;
long double calc(long double x) {
  dp[0][0] = dp[0][1] = 0;
  long double ans0 = 0, ans1 = 0;
  for (int i = 1; i <= n; ++i) {
    dp[i][0] = max(dp[i - 1][0] + data[i] - x, data[i] - x);
    dp[i][1] = min(dp[i - 1][1] + data[i] - x, data[i] - x);
    ans0 = max(ans0, dp[i][0]);
    ans1 = min(ans1, dp[i][1]);
  }
  return max(fabsl(ans0), fabsl(ans1));
}
int main() {
  scanf("%d", &n);
  long double l = 11000, r = -11000;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", data + i);
    l = min(l, (long double)data[i]);
    r = max(r, (long double)data[i]);
  }
  while (r - l > eps) {
    long double mid2 = r * 2 / 3 + l / 3, mid1 = l * 2 / 3 + r / 3;
    if (calc(mid1) > calc(mid2))
      l = mid1;
    else
      r = mid2;
  }
  printf("%f\n", (double)calc((l + r) / 2));
  return 0;
}