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 N = 2e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const double eps = 1e-8;
int n;
double a[N], b[N];
double check(double x) {
  for (int i = 1; i <= n; ++i) b[i] = a[i] - x;
  double sum = 0, ans1 = 0, ans2 = 0;
  for (int i = 1; i <= n; ++i) {
    sum += b[i];
    if (sum > ans1) ans1 = sum;
    if (sum < 1e-10) sum = 0;
  }
  sum = 0;
  for (int i = 1; i <= n; ++i) {
    sum += b[i];
    if (sum < ans2) ans2 = sum;
    if (sum > 1e-10) sum = 0;
  }
  return max(ans1, -ans2);
}
int main() {
  ios_base::sync_with_stdio(0);
  while (scanf("%d", &n) == 1) {
    for (int i = 1; i <= n; ++i) scanf("%lf", a + i);
    double l = -1e10, r = 1e10;
    for (int i = 1; i <= 300; ++i) {
      double ll = (l + r) / 2, rr = (ll + r) / 2;
      if (check(ll) < check(rr))
        r = rr;
      else
        l = ll;
    }
    printf("%.10f\n", check(l));
  }
  return 0;
}