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 mod = 1000000007;
const long long INF = mod * mod;
const long double eps = 1e-6;
const long double pi = acos(-1.0);
int n;
vector<long double> a;
long double calc(long double x) {
  vector<long double> ra(n + 1);
  for (int i = 0; i < n; i++) {
    ra[i + 1] = ra[i] + a[i] - x;
  }
  long double ret = 0;
  long double mi = 0, ma = 0;
  for (int i = 1; i <= n; i++) {
    ret = max(ret, abs(ra[i] - mi));
    ret = max(ret, abs(ma - ra[i]));
    mi = min(mi, ra[i]);
    ma = max(ma, ra[i]);
  }
  return ret;
}
void solve() {
  cout << fixed << setprecision(7);
  cin >> n;
  a.resize(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  long double le = -10000, ri = 10000;
  for (int aa = 0; aa < 100; aa++) {
    long double l = (le * 2 + ri) / 3.0;
    long double r = (le + 2 * ri) / 3.0;
    long double cl = calc(l), cr = calc(r);
    if (cl < cr)
      ri = r;
    else
      le = l;
  }
  cout << calc(le) << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  solve();
  return 0;
}