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;
using lli = long long int;
using ulli = unsigned long long int;
using ld = long double;
int n;
ld a[(int)(2e5 + 5)];
ld f(ld x) {
  ld ans = 0;
  ld temp = 0;
  for (int i = 0; i < n; ++i) {
    temp += a[i] - x;
    if (temp < 0) temp = 0;
    ans = max(ans, temp);
  }
  temp = 0;
  for (int i = 0; i < n; ++i) {
    temp += a[i] - x;
    if (temp > 0) temp = 0;
    ans = max(ans, -temp);
  }
  return ans;
}
int ts(ld l, ld r) {
  if (r - l <= (long double)(1e-12)) {
    cout << setprecision(12) << f(r);
    return 0;
  }
  ld m1 = l + (r - l) / 3;
  ld m2 = r - (r - l) / 3;
  ld f1 = f(m1), f2 = f(m2);
  if (f1 < f2) {
    ts(l, m2);
    return 0;
  }
  if (f2 < f1) {
    ts(m1, r);
    return 0;
  }
  ts(m1, m2);
  return 0;
}
int main() {
  cin >> n;
  ld lmax = -(long double)(10000), lmin = (long double)(10000);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  ts(-(long double)(10000), (long double)(10000));
  return 0;
}