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;
vector<int> v1;
pair<int, int> p1;
double a[1234567];
string s;
int dp[1001];
double max1 = 0.0000000000, max2 = 0.00000000;
double b[200001];
int n;
double check(double b[]) {
  int i;
  double max_upto = 0.0000, overall = 0.0000;
  for (i = 0; i < n; i++) {
    max_upto += b[i];
    if (max_upto > overall) overall = max_upto;
    if (max_upto < 0) max_upto = 0;
  }
  return overall;
}
int main() {
  int t;
  int i, j, k, m;
  scanf("%d", &n);
  for (i = 0; i < n; i++) cin >> a[i];
  double l = -1000000000.0000;
  double h = 1000000000, mid;
  for (j = 0; j < 100; j++) {
    mid = (l + h) / 2;
    for (i = 0; i < n; i++) b[i] = a[i] - mid;
    max1 = check(b);
    for (i = 0; i < n; i++) b[i] *= (-1);
    max2 = check(b);
    if (max1 > max2)
      l = mid;
    else
      h = mid;
  }
  printf("%.12lf\n", max(max1, max2));
  return 0;
}