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 i, j, k, l, n, a[200010];
double t, q, p, o, s[200010];
double ab(double x) {
  if (x < 0) x = -x;
  return x;
}
double cal(double p) {
  int i;
  s[0] = 0;
  for (i = 1; i <= n; i++) s[i] = s[i - 1] + a[i] - p;
  double mi = 0, ma = 0, ansx = -10000000, ansn = 1000000;
  for (i = 1; i <= n; i++) {
    if (s[i] < mi) mi = s[i];
    if (s[i] > ma) ma = s[i];
    if (s[i] - mi > ansx) ansx = s[i] - mi;
    if (s[i] - ma < ansn) ansn = s[i] - ma;
  }
  if (ansx < 0) ansx = -ansx;
  if (ansn < 0) ansn = -ansn;
  if (ansn > ansx) ansx = ansn;
  return (ansx);
}
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  double l = -10000, r = 10000, tl, tr, tm, tm2, m, m2;
  tl = cal(l);
  tr = cal(r);
  while (1) {
    m = (r + l) / 2.0;
    tm = cal(m);
    if (ab(tr - tl) < 0.000001) break;
    m2 = (r + m) / 2.0;
    tm2 = cal(m2);
    if (tm > tm2) {
      l = m;
      tl = tm;
    } else {
      r = m2;
      tr = tm2;
    }
  }
  printf("%.9f\n", tm);
  return 0;
}