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 OnlineJudge = 0;
const int maxn = 200000 + 1912;
const long long INF = 1000LL * 1000 * 1000 * 1000;
int n;
int a[maxn];
long long sum[maxn];
void ReadData() {
  cin >> n;
  for (int _a = 1, _b = n, i = _a; i <= n; i++) cin >> a[i];
  for (int _a = 1, _b = n, i = _a; i <= n; i++) sum[i] = sum[i - 1] + a[i];
}
long double f(long double x) {
  long double mn = 0, mx = 0;
  for (int _a = 1, _b = n, i = _a; i <= n; i++) {
    mn = min(mn, sum[i] - x * i);
    mx = max(mx, sum[i] - x * i);
  }
  return abs(mx - mn);
}
void Process() {
  long double low = (long double)*min_element(a + 1, a + 1 + n),
              high = (long double)*max_element(a + 1, a + 1 + n);
  for (int Try = 0, _a = 120; Try < _a; Try++) {
    long double MidL = (high + low + low) / 3, MidR = (high + high + low) / 3;
    if (f(MidL) < f(MidR))
      high = MidR;
    else
      low = MidL;
  }
  cout << fixed << setprecision(9);
  cout << min(f(low), f(high)) << "\n";
}
int main() {
  std ::ios_base::sync_with_stdio(false);
  cin.tie(0);
  ReadData();
  Process();
  return 0;
}