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 = 5e5 + 256;
const char nxt = '\n';
const double eps = 1e-12;
inline bool Pal(const string& s) {
  return equal(s.begin(), s.end(), s.rbegin());
}
inline int readint() {
  bool neg = false;
  int ret = 0;
  char ch;
  ch = getchar();
  while (true) {
    if (ch == '-') {
      neg = true;
      break;
    }
    if (ch >= '0' && ch <= '9') break;
    ch = getchar();
  }
  if (!neg) ret = ch - '0';
  while (true) {
    ch = getchar();
    if (ch < '0' || ch > '9') break;
    ret = ret * 10 + ch - '0';
  }
  return neg ? -ret : ret;
}
int n;
double arr[MAXN];
double ans;
inline double calc() {
  double ans = 0;
  double sum = 0;
  for (int i = 1; i <= n; ++i) {
    sum += arr[i];
    if (sum > ans) ans = sum;
    if (sum < 0) sum = 0;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout << setprecision(10) << fixed;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> arr[i];
  }
  double l = -1000000.0000, r = 1000000.00000;
  for (int i = 1; i <= 300; ++i) {
    double mid = (l + r) / 2.000;
    for (int j = 1; j <= n; ++j) arr[j] -= mid;
    double cur1 = calc();
    for (int j = 1; j <= n; ++j) arr[j] = -arr[j];
    double cur2 = calc();
    for (int j = 1; j <= n; ++j) arr[j] = -arr[j], arr[j] += mid;
    if (cur1 >= cur2) {
      l = mid;
      ans = cur1;
    } else
      r = mid;
  }
  cout << ans;
  return 0;
}