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 n, a[200000 + 1];
double x[200000 + 1];
double dp1[200000 + 1], dp2[200000 + 1], dp[200000 + 1];
double f(double delta) {
  for (int i = 0; i < n; i++) x[i] = a[i] - delta;
  dp1[0] = x[0];
  for (int i = 1; i < n; i++) {
    dp1[i] = max(dp1[i - 1] + x[i], x[i]);
  }
  dp2[0] = x[0];
  for (int i = 1; i < n; i++) {
    dp2[i] = min(dp2[i - 1] + x[i], x[i]);
  }
  dp[0] = abs(x[0]);
  for (int i = 1; i < n; i++) {
    dp[i] = abs(x[i]);
    dp[i] = max(abs(dp1[i - 1] + x[i]), dp[i]);
    dp[i] = max(abs(dp2[i - 1] + x[i]), dp[i]);
  }
  double ans = dp[0];
  for (int i = 1; i < n; i++) ans = max(ans, dp[i]);
  return ans;
}
int main() {
  while (cin >> n) {
    for (int i = 0; i < n; i++) scanf("%d", &a[i]);
    double lo = -10001, hi = 10001;
    for (int i = 0; i < 100; i++) {
      double m1 = lo + 1.0 * (hi - lo) / 3;
      double m2 = lo + 2.0 * (hi - lo) / 3;
      double x1 = f(m1), x2 = f(m2);
      if (x1 > x2)
        lo = m1;
      else
        hi = m2;
    }
    printf("%.10lf\n", f((lo + hi) / 2.0));
  }
  return 0;
}