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 = 7005;
const double long EPS = 1e-12;
const double long INF = 1e9;
vector<int> ARR, ARR1;
int N;
double long largest1(double long x, vector<int> A) {
  double long max_h = -0, max_s = -INF;
  for (int i = 0; i < N; i++) {
    max_h += A[i] - x;
    max_s = max(max_s, max_h);
    if (max_h < 0) max_h = 0;
  }
  return max_s;
}
double long largest2(double long x, vector<int> A) {
  double long max_h = -0, max_s = -INF;
  for (int i = 0; i < N; i++) {
    max_h += A[i] + x;
    max_s = max(max_s, max_h);
    if (max_h < 0) max_h = 0;
  }
  return max_s;
}
double long sum(double long x) {
  double long s1 = largest1(x, ARR);
  double long s2 = largest2(x, ARR1);
  s1 = max(s1, -s1);
  s2 = max(s2, -s2);
  return max(s1, s2);
}
double long ternary(double long lo, double long hi) {
  while (hi - lo > EPS) {
    double long m1 = lo + (hi - lo) / 3;
    double long m2 = hi - (hi - lo) / 3;
    if (sum(m1) < sum(m2))
      hi = m2;
    else
      lo = m1;
  }
  return hi;
}
int main() {
  scanf("%d", &N);
  for (int i = 0; i < N; i++) {
    int u;
    scanf("%d", &u);
    ARR.push_back(u);
  }
  int MIN = 1e4;
  int MAX = -1e4;
  for (int i = 0; i < N; i++) {
    MIN = min(MIN, ARR[i]);
    MAX = max(MAX, ARR[i]);
  }
  for (int i = 0; i < N; i++) ARR1.push_back(-ARR[i]);
  double long ans = ternary(MIN, MAX);
  ans = sum(ans);
  printf("%Lf", ans);
}