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;
double ar[200009], pr[200009];
int N;
double solve(double x) {
  int i;
  double ans, temp;
  for (i = 1; i <= N; i++) {
    pr[i] = ar[i] - x;
  }
  ans = 0.0;
  temp = 0.0;
  for (i = 1; i <= N; i++) {
    temp += pr[i];
    ans = max(ans, temp);
    if (temp < 0) {
      temp = 0.0;
    }
  }
  for (i = 1; i <= N; i++) {
    pr[i] = -pr[i];
  }
  temp = 0.0;
  for (i = 1; i <= N; i++) {
    temp += pr[i];
    ans = max(ans, temp);
    if (temp < 0) {
      temp = 0.0;
    }
  }
  return ans;
}
int main() {
  int cnt, i;
  double templeft, tempright, z, x, left, right;
  cin >> N;
  for (i = 1; i <= N; i++) cin >> ar[i];
  left = -30000;
  right = 30000;
  cnt = 0;
  while (cnt <= 150) {
    cnt++;
    templeft = left * 2.0 / 3.0 + right / 3.0;
    tempright = right * 2.0 / 3.0 + left / 3.0;
    z = solve(templeft);
    x = solve(tempright);
    if (z > x) {
      left = templeft;
    } else {
      right = tempright;
    }
  }
  printf("%.10lf\n", z);
}