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;
int a[200010];
double temp[200010];
double dp[200010];
double solve(int n, double x) {
  double max = -1000000000;
  for (int i = 0; i < n; i++) temp[i] = (double)a[i] - x;
  for (int i = 0; i < n; i++) {
    if (i == 0)
      dp[i] = temp[i];
    else {
      if (dp[i - 1] > 0)
        dp[i] = dp[i - 1] + temp[i];
      else
        dp[i] = temp[i];
    }
    if (dp[i] > max) max = dp[i];
  }
  for (int i = 0; i < n; i++) temp[i] = -temp[i];
  for (int i = 0; i < n; i++) {
    if (i == 0)
      dp[i] = temp[i];
    else {
      if (dp[i - 1] > 0)
        dp[i] = dp[i - 1] + temp[i];
      else
        dp[i] = temp[i];
    }
    if (dp[i] > max) max = dp[i];
  }
  return max;
}
int main() {
  int i, j, n;
  while (scanf("%d", &n) != EOF) {
    for (i = 0; i < n; i++) scanf("%d", &a[i]);
    double left = -11000, right = 11000;
    int cnt = 0;
    while (left < right && cnt < 100) {
      double ll = (left + (right - left) / 3);
      double rr = (left + (right - left) / 3 * 2);
      double value1 = solve(n, ll);
      double value2 = solve(n, rr);
      if (value1 < value2)
        right = rr;
      else
        left = ll;
      cnt++;
    }
    printf("%.10lf\n", solve(n, left));
  }
  return 0;
}