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 max1, sum, a[200001], b[200001], n;
double f(double y) {
  max1 = 0;
  long long i;
  double sum1 = 0, sum2 = 0;
  for (i = 1; i <= n; i++) a[i] = b[i];
  for (i = 1; i <= n; i++) a[i] -= y;
  for (i = 1; i <= n; i++) {
    sum1 += a[i];
    sum2 += a[i];
    max1 = max(max1, max(abs(sum1), abs(sum2)));
    if (sum2 < 0) sum2 = 0;
    if (sum1 > 0) sum1 = 0;
  }
  return max1;
}
int main() {
  double ri, le, ri1, le1;
  long long i, j;
  cin >> n;
  for (i = 1; i <= n; i++) cin >> b[i];
  le = -10000;
  ri = 10000;
  le1 = le + (ri - le) / 3;
  ri1 = ri - (ri - le) / 3;
  while (1) {
    if (ri - le < 0.00000000001) {
      max1 = f((ri + le) / 2);
      ;
      break;
    } else {
      le1 = le + (ri - le) / 3;
      ri1 = ri - (ri - le) / 3;
      if (f(le1) > f(ri1)) {
        le = le1;
      } else {
        ri = ri1;
      }
    }
  }
  cout.precision(15);
  cout << fixed << max1;
  return 0;
}