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 mod = 1e9 + 7;
int n;
vector<double> v;
double func(double x) {
  double sum = (v[0] - x);
  int now;
  if ((v[0] - x) < 0)
    now = -1;
  else
    now = 1;
  vector<double> a1;
  vector<double> a2;
  for (int i = 1; i < n; i++) {
    if (v[i] - x <= 0) {
      if (now == -1)
        sum += (v[i] - x);
      else {
        now = -1;
        a1.push_back(sum);
        a2.push_back(-sum);
        sum = v[i] - x;
      }
    } else {
      if (now == 1)
        sum += (v[i] - x);
      else {
        a1.push_back(sum);
        a2.push_back(-sum);
        now = 1;
        sum = v[i] - x;
      }
    }
  }
  a1.push_back(sum);
  a2.push_back(-sum);
  double ans = 0;
  sum = 0;
  for (int i = 0; i < a1.size(); i++) {
    sum += a1[i];
    ans = max(sum, ans);
    if (sum < 0) sum = 0;
  }
  sum = 0;
  for (int i = 0; i < a2.size(); i++) {
    sum += a2[i];
    ans = max(ans, sum);
    if (sum < 0) sum = 0;
  }
  return ans;
}
void solve() {
  cin >> n;
  v = vector<double>(n);
  for (int i = 0; i < n; i++) cin >> v[i];
  double lo = -1e4 - 10;
  double hi = 1e4 + 10;
  int cnt = 0;
  while (true) {
    double m1 = lo + (hi - lo) / 3;
    double m2 = hi - (hi - lo) / 3;
    double f1 = func(m1);
    double f2 = func(m2);
    if (f1 > f2) {
      lo = m1;
    } else
      hi = m2;
    cnt++;
    if (cnt == 100) break;
  }
  printf("%0.9f\n", func(lo));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int tests = 1;
  while (tests--) {
    solve();
  }
}