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;
long double f(long double val);
long double vet[200010], aux[200010], l, r;
int n;
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> vet[i];
  }
  l = -10000;
  r = 10000;
  while (abs(r - l) > 0.000000000001) {
    long double lT = l + (r - l) / 3;
    long double rT = r - (r - l) / 3;
    if (f(lT) > f(rT)) {
      l = lT;
    } else {
      r = rT;
    }
  }
  cout << fixed;
  cout << setprecision(12) << f(l) << endl;
  return 0;
}
long double f(long double val) {
  for (int i = 0; i < n; i++) {
    aux[i] = vet[i] + val;
  }
  long double pPre, nPre, pMax, nMin;
  if (aux[0] >= 0) {
    pMax = aux[0];
    pPre = aux[0];
    nPre = 0;
    nMin = 0;
  } else {
    pMax = 0;
    pPre = 0;
    nPre = aux[0];
    nMin = aux[0];
  }
  for (int i = 1; i < n; i++) {
    if (aux[i] >= 0) {
      pPre += aux[i];
      nPre = min(nPre + aux[i], (long double)0);
      if (pPre > pMax) {
        pMax = pPre;
      }
    } else {
      nPre += aux[i];
      pPre = max(pPre + aux[i], (long double)0);
      if (nPre < nMin) {
        nMin = nPre;
      }
    }
  }
  nMin = abs(nMin);
  return max(nMin, pMax);
}