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 long long N = 2e5 + 5;
const long long mod = 1e9 + 7;
long long n;
long double l, r, ans = 1e18, arr[N], sum[N];
long double func(long double val) {
  long double mn = 0, mx = 0, fin = -1e12;
  for (long long i = 1; i <= n; i++) {
    sum[i] = sum[i - 1] + arr[i] - val;
    fin = max(fin, max(sum[i] - mn, mx - sum[i]));
    mx = max(mx, sum[i]);
    mn = min(mn, sum[i]);
  }
  return fin;
}
int main() {
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> arr[i];
  r = 1e6, l = -1e6;
  ans = func(0);
  for (long long i = 1; i <= 200; i++) {
    long double mid1 = l + (r - l) / 3;
    long double mid2 = r - (r - l) / 3;
    long double x = func(mid1), y = func(mid2);
    ans = min(ans, min(x, y));
    if (x < y)
      r = mid2;
    else
      l = mid1;
  }
  cout << fixed << setprecision(18) << ans;
}