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 N = 100000 * 2 + 10;
int a[N];
int n;
long double majmo(long double x) {
  long double b[n];
  for (int i = 0; i < n; i++) {
    b[i] = a[i] - x;
  }
  long double dp[N];
  long double dp1[N];
  dp[0] = 0;
  dp1[0] = 0;
  for (int i = 0; i < n; i++) {
    if (dp1[i] + b[i] < 0) {
      dp1[i + 1] = 0;
    } else {
      dp1[i + 1] = dp1[i] + b[i];
    }
    dp[i + 1] = max(dp[i], dp1[i + 1]);
  }
  long double h = dp[n];
  for (int i = 0; i < n; i++) {
    b[i] = -b[i];
    dp[i] = 0;
    dp1[i] = 0;
  }
  dp[n] = 0;
  dp1[n] = 0;
  for (int i = 0; i < n; i++) {
    if (dp1[i] + b[i] < 0) {
      dp1[i + 1] = 0;
    } else {
      dp1[i + 1] = dp1[i] + b[i];
    }
    dp[i + 1] = max(dp[i], dp1[i + 1]);
  }
  return max(dp[n], h);
}
int main() {
  cin >> n;
  long double mi = 100000, ma = -1000000;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] < mi) {
      mi = a[i];
    }
    if (a[i] > ma) {
      ma = a[i];
    }
  }
  long double mid1, mid2;
  for (int i = 0; i < 100; i++) {
    mid1 = (2 * mi + ma) / 3;
    mid2 = (2 * ma + mi) / 3;
    if (majmo(mid1) < majmo(mid2)) {
      ma = mid2;
    } else {
      mi = mid1;
    }
  }
  cout << fixed << setprecision(20) << majmo(mid1);
}