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 a[200010], sum, mx, mn;
int n, x;
char c;
bool b;
double f(double x) {
  sum = 0, mx = 0, mn = 0;
  for (int i = 0; i < n; i++) {
    sum += a[i] - x;
    if (mx < sum) mx = sum;
    if (mn > sum) mn = sum;
  }
  return (mx - mn);
}
int qin() {
  b = false, x = 0;
  c = getchar();
  if (c == '-') {
    b = true;
    c = '0';
  }
  while (c != ' ') {
    x = x * 10 + (c - '0');
    c = getchar();
  }
  if (b) return -x;
  return x;
}
int qin2() {
  b = false, x = 0;
  c = getchar();
  if (c == '-') {
    b = true;
    c = '0';
  } else
    x = c - '0';
  while ((c = getchar()) != EOF && c >= '0' && c <= '9') x = x * 10 + (c - '0');
  if (b) return -x;
  return x;
}
int main() {
  n = qin2();
  for (int i = 0; i < n - 1; i++) a[i] = qin();
  a[n - 1] = qin2();
  double l = -10001, r = 10001, ml, mr, z1, z2;
  int x = 0;
  while (r - l >= 0.00000000001) {
    ml = (l + r) / 2.0;
    mr = (ml + r) / 2.0;
    z1 = f(mr), z2 = f(ml);
    if (z1 < z2)
      l = ml;
    else
      r = mr;
    x++;
  }
  l = (l + r) / 2.0;
  printf("%.15lf", f(l));
  return 0;
}