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 = 200000 + 10;
double a[MAXN];
double dp[MAXN];
double f[MAXN];
int n;
double cal(double x) {
  double res = 0;
  f[1] = dp[1] = a[1] - x;
  res = fabs(dp[1]);
  for (int i = 2; i <= n; i++) {
    double tmp = a[i] - x;
    dp[i] = max(tmp, dp[i - 1] + tmp);
    f[i] = min(tmp, tmp + f[i - 1]);
    res = max(res, fabs(dp[i]));
    res = max(res, fabs(f[i]));
  }
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lf", a + i);
  double L = -MAXN, R = MAXN;
  for (int i = 1; i <= 300; i++) {
    double m1 = L + (R - L) / 3;
    double m2 = R - (R - L) / 2;
    if (cal(m1) < cal(m2)) {
      R = m2;
    } else {
      L = m1;
    }
  }
  printf("%.12f\n", cal(L));
}