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 target("avx,avx2,fma")
using namespace std;
template <typename T>
int len(const T& a) {
  return a.size();
}
using ll = long long;
constexpr int N = 2e5 + 10;
constexpr long double EPS = 1e-7;
int n;
int a[N];
long double Check(long double x) {
  static long double new_a[N];
  long double bestmin = 1e8, curmin = 0;
  long double bestmax = -1e8, curmax = 0;
  for (int i = 0; i < n; ++i) {
    new_a[i] = (long double)a[i] - x;
    curmax = max((long double)0, curmax + new_a[i]);
    bestmax = max(bestmax, curmax);
    curmin = min((long double)0, curmin + new_a[i]);
    bestmin = min(bestmin, curmin);
  }
  return max(abs(bestmin), abs(bestmax));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
  long double l = -1e4 - 10, r = 1e4 + 10;
  for (int step = 0; step < 500; ++step) {
    long double m1 = l + (r - l) / 3.0;
    long double m2 = r - (r - l) / 3.0;
    if (Check(m1) - Check(m2) > EPS) {
      l = m1;
    } else {
      r = m2;
    }
  }
  printf("%.6lf", (double)Check(l));
  return 0;
}