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 MOD = 1e9 + 7;
const int maxn = 300009;
const int maxm = 300009;
const double pi = acos(-1.0);
const double eps = 1e-15;
double data[maxn], sum, maxx, minx;
double tmp[maxn];
double f2(double x, int n) {
  for (int i = 1; i <= n; i++) tmp[i] = data[i] - x;
  double tmpp = tmp[1], tmpn = tmp[1];
  double ans = tmpp, ans2 = tmpn;
  for (int i = 2; i <= n; i++) {
    if (tmpp > 0)
      tmpp += tmp[i];
    else
      tmpp = tmp[i];
    ans = max(tmpp, ans);
    if (tmpn < 0)
      tmpn += tmp[i];
    else
      tmpn = tmp[i];
    ans2 = min(tmpn, ans2);
  }
  return max(abs(ans), abs(ans2));
}
double sanfen(int n) {
  double l = minx, r = maxx, mid = l + (r - l) / 3., midd = r - (r - l) / 3.;
  int cnt = 100;
  while (fabs(r - l) > eps && cnt--) {
    double fmid = f2(mid, n), fmidd = f2(midd, n);
    if (fmid < fmidd)
      r = midd;
    else
      l = mid;
    mid = l + (r - l) / 3., midd = r - (r - l) / 3.;
  }
  return f2((l + r) / 2., n);
}
int main() {
  int n;
  while (~scanf("%d", &n)) {
    maxx = -1000000.;
    minx = 1000000.;
    sum = 0.0;
    for (int i = 1; i <= n; i++) scanf("%lf", &data[i]), sum += data[i];
    if (n != 1)
      printf("%.15f\n", sanfen(n));
    else
      printf("%.15lf\n", 0.0);
  }
  return 0;
}