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 maxn = 2e5 + 2;
const long long inf = 1e18 + 43;
const int mod = 1e9 + 7;
const int maxlog = 20;
const long double pi = 3.1415926535897932384626433832795028841971693993751;
long double a[maxn];
int n;
long double get(long double x) {
  long double mini = 0, maxi = 0;
  long double sum = 0;
  for (int i = 0; i < n; i++) {
    long double b = a[i] - x;
    sum += b;
    mini = min(mini, sum);
    maxi = max(maxi, sum);
  }
  return maxi - mini;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout.precision(20);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  long double l = -1e9, r = 1e9;
  int t = 300;
  while (t--) {
    long double mid1 = l + (r - l) / 3;
    long double mid2 = l + 2 * (r - l) / 3;
    if (get(mid1) >= get(mid2))
      l = mid1;
    else
      r = mid2;
  }
  cout << get(l);
  return 0;
}