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 MAXN = (int)2e5 + 10;
const int INF = (int)0x7f7f7f7f;
int n, a[MAXN];
double b[MAXN], s[MAXN];
double MCS() {
  s[0] = 0;
  for (int i = 1; i <= n; ++i) s[i] = s[i - 1] + b[i];
  double m = 0;
  double ans = -INF;
  for (int i = 1; i <= n; ++i) {
    ans = max(ans, s[i] - m);
    m = min(m, s[i]);
  }
  return ans;
}
void Cal(double x, double &p, double &q) {
  for (int i = 1; i <= n; ++i) b[i] = a[i] - x;
  p = MCS();
  for (int i = 1; i <= n; ++i) b[i] = x - a[i];
  q = MCS();
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  double l = *min_element(a + 1, a + 1 + n);
  double r = *max_element(a + 1, a + 1 + n);
  for (int k = 0; k <= 500; ++k) {
    double m = (l + r) / 2;
    double x, y;
    Cal(m, x, y);
    if (x > y)
      l = m;
    else
      r = m;
  }
  double x, y, ans;
  Cal(l, x, y);
  ans = max(x, y);
  Cal(r, x, y);
  ans = max(ans, max(x, y));
  cout << fixed << setprecision(10) << ans;
  return 0;
}