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 INF = 2147483647, MOD = 1000 * 1000 * 1000 + 7;
const double eps = 1e-8;
void Error(string err) {
  cout << err;
  cerr << "_" << err << "_";
  exit(0);
}
int gcd(int a, int b) { return (b == 0) ? a : gcd(b, a % b); }
const int sz = 222 * 1000;
int arr[sz], arr2[sz];
double sum[sz];
int n;
double f(double x) {
  double ret = 0;
  double sm = 0;
  for (int i = 0; i < (n); i++) {
    if (sm + arr[i] - x > arr[i] - x)
      sm += arr[i] - x;
    else
      sm = arr[i] - x;
    ret = max(ret, abs(sm));
  }
  sm = 0;
  for (int i = 0; i < (n); i++) {
    if (sm + arr[i] - x < arr[i] - x)
      sm += arr[i] - x;
    else
      sm = arr[i] - x;
    ret = max(ret, abs(sm));
  }
  return ret;
}
int main() {
  cin >> n;
  for (int i = 0; i < (n); i++) cin >> arr[i];
  sum[0] = arr[0];
  for (int i = 1; i <= (n - 1); i++) sum[i] = sum[i - 1] + arr[i];
  for (int i = 0; i < (n); i++) arr2[i] = arr[i];
  sort(arr2, arr2 + n);
  double a = arr2[n / 2];
  double wa = f(a);
  double inc = (arr2[n - 1] - arr2[0]);
  while (true) {
    wa = f(a);
    double nwai = f(a + inc), nwad = f(a - inc);
    if (wa > nwai && wa < nwad)
      a += inc;
    else if (wa > nwad && wa < nwai)
      a -= inc;
    else
      inc /= 2;
    double x = wa - nwai;
    double y = wa - nwad;
    double z = nwai - nwad;
    x = abs(x) + abs(y) + abs(z);
    if (x < 1e-9) break;
  }
  cout.setf(ios::fixed);
  cout.precision(8);
  cout << wa << endl;
  return 0;
}