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;
inline long long int GCD(long long int x, long long int y) {
  if (x < y) swap(x, y);
  if (x == 0) return y;
  if (y == 0) return x;
  return GCD(x % y, y);
}
long long int phi(long long int n) {
  long long int result = n;
  for (long long int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      while (n % i == 0) n /= i;
      result -= result / i;
    }
  }
  if (n > 1) result -= result / n;
  return result;
}
long long int power(long long int x, long long int n, long long int mod) {
  long long int res = 1;
  x %= mod;
  while (n) {
    if (n & 1) {
      res = ((res * x) % mod + mod) % mod;
    }
    x = ((x * x) % mod + mod) % mod;
    n >>= 1;
  }
  return res;
}
const long double eps = 1e-12;
const int MAXN = 200005;
long double N, A[MAXN], Al[MAXN];
long double process(long double x) {
  for (long long int i = 1; i <= N; i++) {
    Al[i] = A[i] - x;
  }
  long double sum = 0;
  long double mx = -1e18 + 9;
  for (long long int i = 1; i <= N; i++) {
    sum += Al[i];
    if (sum < 0) sum = 0;
    mx = max(mx, sum);
  }
  for (long long int i = 1; i <= N; i++) {
    Al[i] = -Al[i];
  }
  sum = 0;
  for (long long int i = 1; i <= N; i++) {
    sum += Al[i];
    if (sum < 0) sum = 0;
    mx = max(mx, sum);
  }
  return mx;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> N;
  for (long long int i = 1; i <= N; i++) {
    cin >> A[i];
  }
  long double l = -10000.0, r = 10000.0;
  while (r - l > eps) {
    long double mid1 = l + (long double)(r - l) / (long double)3.0;
    long double mid2 = r - (long double)(r - l) / (long double)3.0;
    if (process(mid1) < process(mid2)) {
      r = mid2;
    } else {
      l = mid1;
    }
  }
  long double ans = process(l);
  cout << setprecision(10) << fixed << ans << endl;
}