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[233333], n;
double tA[233333], dp[233333], mr[233333];
int dcmp(double x) {
  if (x < -1e-11)
    return -1;
  else if (x > 1e-11)
    return 1;
  else
    return 0;
}
double calc(double x) {
  for (int i = 0; i < n; ++i) {
    tA[i] = A[i] - x;
  }
  dp[0] = mr[0] = max(0.0, tA[0]);
  for (int i = 1; i < n; ++i) {
    mr[i] = max(0.0, tA[i] + mr[i - 1]);
    dp[i] = max(dp[i - 1], mr[i]);
  }
  double y1 = dp[n - 1];
  for (int i = 0; i < n; ++i) {
    tA[i] = x - A[i];
  }
  dp[0] = mr[0] = max(0.0, tA[0]);
  for (int i = 1; i < n; ++i) {
    mr[i] = max(0.0, tA[i] + mr[i - 1]);
    dp[i] = max(dp[i - 1], mr[i]);
  }
  double y2 = dp[n - 1];
  return max(y1, y2);
}
int main() {
  while (~scanf("%d", &n)) {
    for (int i = 0; i < n; ++i) {
      scanf("%d", &A[i]);
    }
    double L = -2e4, R = 2e4;
    while (dcmp(L - R)) {
      double m1 = L * 2 / 3 + R / 3;
      double m2 = L / 3 + R * 2 / 3;
      if (calc(m1) < calc(m2))
        R = m2;
      else
        L = m1;
    }
    printf("%.9f\n", calc((L + R) / 2.0));
  }
  return 0;
}