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;
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
int32_t main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  long long int n;
  cin >> n;
  long double a[n + 1], ans = 10000000000000000.00, l = -10000.00, r = 10000.00;
  for (long long int i = 1; i <= n; i++) cin >> a[i];
  for (long long int i = 0; i <= 100; i++) {
    long double m1 = l + (r - l) / ((long double)3);
    long double m2 = r - (r - l) / ((long double)3);
    long double t1 = 0.00, t2 = 0.00, s1 = -100000, s2 = -100000;
    for (long long int j = 1; j <= n; j++) {
      t1 += (long double)(a[j] - m1);
      t2 += (long double)(a[j] - m2);
      s1 = (long double)max(s1, t1);
      s2 = (long double)max(s2, t2);
      if (t1 < 0) t1 = 0.00;
      if (t2 < 0) t2 = 0.00;
    }
    t1 = t2 = 0.00;
    for (long long int j = 1; j <= n; j++) {
      t1 += (long double)(a[j] - m1);
      t2 += (long double)(a[j] - m2);
      s1 = (long double)max(s1, abs(t1));
      s2 = (long double)max(s2, abs(t2));
      if (t1 > 0) t1 = 0.00;
      if (t2 > 0) t2 = 0.00;
    }
    if (s1 < s2)
      r = m2;
    else
      l = m1;
    ans = min({ans, s1, s2});
  }
  cout << fixed << setprecision(10) << ans;
}