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;
struct __s {
  __s() {
    if (1) {
      ios_base::Init i;
      cin.sync_with_stdio(0);
      cin.tie(0);
    }
  }
  ~__s() {
    if (!1)
      fprintf(stderr, "Execution time: %.3lf s.\n",
              (double)clock() / CLOCKS_PER_SEC);
    long long n;
    cin >> n;
  }
} __S;
long long n;
double a[211111];
double mx() {
  double ans = a[0];
  double sum = 0;
  double min_sum = 0;
  for (long long r = 0; r < (long long)(n); r++) {
    sum += a[r];
    ans = max(ans, sum - min_sum);
    min_sum = min(min_sum, sum);
  }
  return ans;
}
int main(void) {
  cin >> n;
  for (long long i = 0; i < (long long)(n); i++) {
    long long x;
    cin >> x;
    a[i] = x;
  }
  double l = 2e4;
  double r = -2e4;
  for (long long _ = 0; _ < (long long)(60); _++) {
    double x = (l + r) / 2.0;
    for (long long i = 0; i < (long long)(n); i++) {
      a[i] += x;
    }
    double mx1 = mx();
    for (long long i = 0; i < (long long)(n); i++) {
      a[i] *= -1;
    }
    double mx2 = mx();
    for (long long i = 0; i < (long long)(n); i++) {
      a[i] *= -1;
      a[i] -= x;
    }
    if (mx1 >= mx2) {
      l = x;
    } else {
      r = x;
    }
  }
  double x = l;
  for (long long i = 0; i < (long long)(n); i++) {
    a[i] += x;
  }
  double mx1 = mx();
  for (long long i = 0; i < (long long)(n); i++) {
    a[i] *= -1;
  }
  double mx2 = mx();
  cout << fixed << setprecision(10) << max(mx1, mx2) << '\n';
  return 0;
}