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>
#pragma GCC optimize("", on)
using namespace std;
const long long maxn = 5e5 + 13;
const long long INF = 2e9 + 13;
const long long MOD = 1e9 + 7;
double a[maxn];
double f(double x, long long n) {
  vector<double> pref(n), b(n);
  double minS = INF, maxS = -INF;
  pref[0] = a[1] - x;
  for (long long i = 1; i <= n; ++i)
    b[i - 1] = a[i] - x, minS = min(a[i] - x, minS), maxS = max(maxS, a[i] - x);
  for (long long i = 1; i < n; ++i) pref[i] = pref[i - 1] + b[i];
  double minl = 0;
  for (long long i = 0; i < n; ++i) {
    maxS = max(maxS, pref[i] - minl);
    minl = min(pref[i], minl);
  }
  minl = 0;
  for (long long i = 0; i < n; ++i) {
    minS = min(minS, pref[i] - minl);
    minl = max(minl, pref[i]);
  }
  return max(abs(minS), abs(maxS));
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  double maxe = 0;
  for (long long i = 1; i <= n; ++i) cin >> a[i], maxe = max(maxe, abs(a[i]));
  double l = -maxe - 1, r = maxe + 1;
  for (long long i = 1; i <= 100; ++i) {
    double x1 = l + (r - l) / 3;
    double x2 = l + (2 * (r - l)) / 3;
    double f1 = f(x1, n);
    double f2 = f(x2, n);
    if (f1 > f2)
      l = x1;
    else
      r = x2;
  }
  cout << fixed << setprecision(10) << f(l, n) << "\n";
  return 0;
}