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 double eps = 1e-12;
const long long MOD = 1e9 + 7;
int dx[4] = {1, 0, 0, -1};
int dy[4] = {0, 1, -1, 0};
void fail() {
  cout << "-1";
  exit(0);
}
void win() {
  cout << "YES";
  exit(0);
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long double p[300300], a[300300];
int n;
long double f(long double x) {
  p[0] = a[0] - x;
  for (int i = 1; i < n; i++) p[i] = p[i - 1] + a[i] - x;
  long double ans1 = -2e12, ans2 = 2e12;
  long double mn = 2e9 + 41;
  for (int i = 0; i < n; i++) {
    ans1 = max(p[i] - mn, max(p[i], ans1));
    mn = min(mn, p[i]);
  }
  mn = -2e9 - 41;
  for (int i = 0; i < n; i++) {
    ans2 = min(p[i] - mn, min(p[i], ans2));
    mn = max(mn, p[i]);
  }
  return max(fabs(ans1), fabs(ans2));
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  long double l = -1e5, r = 1e5;
  for (int q = 0; q < 100; q++) {
    long double m1 = l + (r - l) / 3, m2 = r - (r - l) / 3;
    if (f(m1) < f(m2))
      r = m2;
    else
      l = m1;
  }
  cout.precision(10);
  cout << fixed << f(l);
  return 0;
}