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 long MulModL(long long B, long long P, long long M) {
  long long R = 0;
  while (P > 0) {
    if ((P & 1ll) == 1) {
      R = (R + B);
      if (R >= M) R -= M;
    }
    P >>= 1ll;
    B = (B + B);
    if (B >= M) B -= M;
  }
  return R;
}
long long MulModD(long long B, long long P, long long M) {
  long long I = ((long double)B * (long double)P / (long double)M);
  long long R = B * P - M * I;
  R = (R % M + M) % M;
  return R;
}
long long BigMod(long long B, long long P, long long M) {
  long long R = 1;
  while (P > 0) {
    if (P % 2 == 1) {
      R = (R * B) % M;
    }
    P /= 2;
    B = (B * B) % M;
  }
  return R;
}
template <class T1>
void deb(T1 e1) {
  cout << e1 << endl;
}
template <class T1, class T2>
void deb(T1 e1, T2 e2) {
  cout << e1 << " " << e2 << endl;
}
template <class T1, class T2, class T3>
void deb(T1 e1, T2 e2, T3 e3) {
  cout << e1 << " " << e2 << " " << e3 << endl;
}
template <class T1, class T2, class T3, class T4>
void deb(T1 e1, T2 e2, T3 e3, T4 e4) {
  cout << e1 << " " << e2 << " " << e3 << " " << e4 << endl;
}
template <class T1, class T2, class T3, class T4, class T5>
void deb(T1 e1, T2 e2, T3 e3, T4 e4, T5 e5) {
  cout << e1 << " " << e2 << " " << e3 << " " << e4 << " " << e5 << endl;
}
template <class T1, class T2, class T3, class T4, class T5, class T6>
void deb(T1 e1, T2 e2, T3 e3, T4 e4, T5 e5, T6 e6) {
  cout << e1 << " " << e2 << " " << e3 << " " << e4 << " " << e5 << " " << e6
       << endl;
}
const int MAX = 200009;
double val[MAX];
int n;
double getRet(double m) {
  double sum = 0;
  double res = 0;
  for (int i = 1; i <= n; i++) {
    sum += (val[i] - m);
    res = max(res, fabs(val[i] - m));
  }
  res = max(res, fabs(sum));
  sum = 0;
  double s = 1.0;
  for (int i = 1; i <= n; i++) {
    sum += (val[i] - m) * s;
    if (sum < 0.0) {
      sum = 0;
    }
    res = max(res, sum);
  }
  s = -1.0;
  sum = 0;
  for (int i = 1; i <= n; i++) {
    sum += (val[i] - m) * s;
    if (sum < 0.0) {
      sum = 0;
    }
    res = max(res, sum);
  }
  return res;
}
int main() {
  scanf("%d", &n);
  int v;
  double sum = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &v);
    val[i] = v;
  }
  double res = 0;
  int cnt = 500;
  double lo = -2.0e10, hi = 2.0e10, m1, m2;
  while (cnt--) {
    m1 = (2.0 * lo + hi) / 3.0;
    m2 = (lo + hi * 2.0) / 3.0;
    double r1 = getRet(m1), r2 = getRet(m2);
    if (r1 < r2) {
      res = r1;
      hi = m2;
    } else {
      res = r2;
      lo = m1;
    }
  }
  printf("%.12lf", res);
  return 0;
}