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;
const int maxn = 200000 + 10;
const int INF = 10000 + 10;
const double eps = 1e-11;
int n, k = 100;
double a[maxn], dp1[maxn], dp2[maxn];
double get_max(double x, double y) { return x > y ? x : y; }
double get_min(double x, double y) { return x < y ? x : y; }
double f(double x) {
  int i;
  double max = 0, b[maxn];
  for (i = 0; i < n; i++) b[i] = a[i] - x;
  dp1[0] = b[0];
  for (i = 1; i < n; i++) dp1[i] = get_max(dp1[i - 1] + b[i], b[i]);
  dp2[0] = b[0];
  for (i = 1; i < n; i++) dp2[i] = get_min(dp2[i - 1] + b[i], b[i]);
  for (i = 0; i < n; i++) {
    max = get_max(max, abs(dp1[i]));
    max = get_max(max, abs(dp2[i]));
  }
  return max;
}
double divide(double &l, double &r) {
  double mid, mmid, fmid, fmmid;
  if (r - l < eps) return (l + r) / 2;
  mid = (r + 2 * l) / 3;
  mmid = (l + 2 * r) / 3;
  fmid = f(mid);
  fmmid = f(mmid);
  if (fmid <= fmmid)
    r = mmid;
  else if (fmid > fmmid)
    l = mid;
  return divide(l, r);
}
int main() {
  int i;
  double ans, max = -INF, min = INF;
  scanf("%d", &n);
  for (i = 0; i < n; i++) scanf("%lf", &a[i]);
  for (i = 0; i < n; i++) {
    if (a[i] > max) max = a[i];
    if (a[i] < min) min = a[i];
  }
  while (k--) {
    divide(min, max);
  }
  ans = divide(min, max);
  printf("%.6lf\n", f(ans));
  return 0;
}