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;
#pragma comment(linker, "/STACK:102400000,102400000")
void RI(int& x) {
  x = 0;
  char c = getchar();
  while (c == ' ' || c == '\n') c = getchar();
  bool flag = 1;
  if (c == '-') {
    flag = 0;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  if (!flag) x = -x;
}
void RII(int& x, int& y) { RI(x), RI(y); }
void RIII(int& x, int& y, int& z) { RI(x), RI(y), RI(z); }
const int maxn = 1e5 + 1000;
const double eps = 1e-11;
int n;
int a[maxn << 1];
double b[maxn << 1];
double cal(double x) {
  double res = 0;
  double now = 0;
  for (int i = 0; i < n; i++) b[i] = a[i] * 1.0 - x;
  for (int i = 0; i < n; i++) {
    now += b[i];
    if (now < 0) now = 0;
    res = max(res, now);
    b[i] = -b[i];
  }
  now = 0;
  for (int i = 0; i < n; i++) {
    now += b[i];
    if (now < 0) now = 0;
    res = max(res, now);
  }
  return res;
}
int main() {
  RI(n);
  for (int i = 0; i < n; i++) RI(a[i]);
  double tmp = 0;
  double l = -10000, r = 10000, mid1, mid2;
  double ans1, ans2;
  int t = 100;
  while (t--) {
    mid1 = l + (r - l) / 3;
    mid2 = r - (r - l) / 3;
    ans1 = cal(mid1);
    ans2 = cal(mid2);
    if (ans1 > ans2)
      l = mid1;
    else
      r = mid2;
  }
  printf("%.6f\n", cal(l));
  return 0;
}