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 iinf = 0x7fffffff;
const long long linf = ~(1LL << 63);
const int maxn = 2e5 + 5;
int n;
double a[maxn];
double b[maxn];
double max_con_seq(double* seq, int len) {
  double ans = -linf;
  double now = 0;
  for (int i = 0; i < len; i++) {
    now += seq[i];
    if (ans < now) ans = now;
    if (now < 0) now = 0;
  }
  return ans;
}
double wrapper(double x) {
  double ans = -linf;
  for (int i = 0; i < n; i++) b[i] = a[i] - x;
  ans = max_con_seq(b, n);
  for (int i = 0; i < n; i++) b[i] = x - a[i];
  ans = max(ans, max_con_seq(b, n));
  return ans;
}
void read() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
}
void solve() {
  double l = -20050, r = 20050;
  double ans = wrapper((l + r) / 2);
  double last_ans = -linf;
  while (fabs(l - r) > 1e-11) {
    double lm = (r + 2 * l) / 3;
    double rm = (2 * r + l) / 3;
    double lv = wrapper(lm), rv = wrapper(rm);
    last_ans = ans;
    if (lv <= rv) {
      r = rm;
      ans = lv;
    } else {
      l = lm;
      ans = rv;
    }
  }
  cout << fixed << setprecision(10) << ans << endl;
}
int main() {
  ios::sync_with_stdio(0);
  read();
  solve();
  return 0;
}