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 dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
int dx8[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dy8[] = {1, -1, -1, 0, 1, -1, 0, 1};
int kx8[] = {1, 1, 2, 2, -1, -1, -2, -2};
int ky8[] = {2, -2, 1, -1, 2, -2, 1, -1};
long long bigmod(long long a, long long b, long long c) {
  if (b == 0) return 1 % c;
  long long x = bigmod(a, b / 2, c);
  x = x * x % c;
  if (b % 2 == 1) x = x * a % c;
  return x;
}
long long mod_inverse(long long a, long long mod) {
  return bigmod(a, mod - 2, mod);
}
double a[2 * 100005];
double b[2 * 100005];
double fun(int n) {
  double mx = -1e15;
  double sum = 0;
  int ptr = 0;
  for (int i = 0; i < (n); i++) {
    sum += b[i];
    while (sum < 0 && ptr <= i) {
      sum -= b[ptr++];
    }
    mx = max(sum, mx);
  }
  return mx;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n;
  for (int i = 0; i < (n); i++) cin >> a[i];
  double lw = -100000, hi = 100000, ans = 0, mid, midd;
  double res = 1e15;
  for (int i = 0; i < (200); i++) {
    mid = (lw + lw + hi) / 3.0;
    midd = (lw + hi + hi) / 3.0;
    for (int i = 0; i < (n); i++) {
      b[i] = a[i] - mid;
    }
    double d1 = fun(n);
    for (int i = 0; i < (n); i++) {
      b[i] *= -1.0;
    }
    double d2 = fun(n);
    double mx = max(d1, d2);
    for (int i = 0; i < (n); i++) {
      b[i] = a[i] - midd;
    }
    d1 = fun(n);
    for (int i = 0; i < (n); i++) {
      b[i] *= -1.0;
    }
    d2 = fun(n);
    double mxx = max(d1, d2);
    if (mx < mxx) {
      hi = midd;
      ans = mid;
      res = mx;
    } else {
      lw = mid;
      ans = midd;
      res = mxx;
    }
  }
  cout << fixed << setprecision(10) << res << endl;
  return 0;
}