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 = 200010;
const int inf = maxn;
int n;
int a[maxn];
int main() {
  cin >> n;
  int maxa = -inf, mina = inf;
  for (int i = (1); i <= (n); i++) {
    scanf("%d", &a[i]);
    maxa = ((maxa) > (a[i]) ? (maxa) : (a[i]));
    mina = ((mina) > (a[i]) ? (a[i]) : (mina));
  }
  double l = mina, r = maxa;
  double mid1 = 0, mid2 = 10, ans1, ans2;
  double maxdp, mindp, maxp, minp;
  for (int ii = (0); ii <= (200); ii++) {
    mid1 = l + (r - l) / 3;
    mid2 = r - (r - l) / 3;
    maxdp = mindp = maxp = minp = 0;
    for (int i = (1); i <= (n); i++) {
      maxdp = (maxdp + a[i] - mid1) < 0 ? 0 : maxdp + a[i] - mid1;
      maxp = ((maxp) > (maxdp) ? (maxp) : (maxdp));
      mindp = (mindp + a[i] - mid1) > 0 ? 0 : mindp + a[i] - mid1;
      minp = ((minp) > (mindp) ? (mindp) : (minp));
    }
    ans1 = max(fabs(maxp), fabs(minp));
    maxdp = mindp = maxp = minp = 0;
    for (int i = (1); i <= (n); i++) {
      maxdp = (maxdp + a[i] - mid2) < 0 ? 0 : maxdp + a[i] - mid2;
      maxp = ((maxp) > (maxdp) ? (maxp) : (maxdp));
      mindp = (mindp + a[i] - mid2) > 0 ? 0 : mindp + a[i] - mid2;
      minp = ((minp) > (mindp) ? (mindp) : (minp));
    }
    ans2 = max(fabs(maxp), fabs(minp));
    if (ans1 > ans2)
      l = mid1;
    else
      r = mid2;
  }
  printf("%.10lf\n", ans2);
  return 0;
}