三分 + 最大区间和

You are given a sequence of n integers a₁, a₂, ..., a_n.

Determine a real number x such that the weakness of the sequence a₁ - x, a₂ - x, ..., a_n - 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, ..., a_n (|a_i| ≤ 10 000).

Output

Output a real number denoting the minimum possible weakness of a₁ - x, a₂ - x, ..., a_n - x. Your answer will be considered correct if its relative or absolute error doesn't exceed 10^- 6.

Example

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.

题意 ： 寻找一个X ，让数组中的全部元素都减去 x ,找区间和最大的情况，所求答案是所有可能情况下最小的答案。

思路分析 ： 当 X 很大时，答案也会是一个很大的值，当 X 很小时，答案也会是一个很大的值，所以，只有当 X 适当时才会有最小的答案，那么这里显然就是三分，图像是一个凹函数，三分判断的条件就是一个求一个最大区间和就可以，因为区间绝对值和最大，它产生的情况有两种，一种区间的和是正值最大，一种区间的和是负值，让它最小，则绝对值会大。

代码示例 ：

#define ll long long
const int maxn = 2e5+5;
const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;

double arr[maxn];
double pp[maxn];
int n;
double dp[maxn], dp2[maxn];
double ans = 1.0*inf;

double abc(double x){
    return x < 0?-x:x;
}

double fun(double x){
    for(int i = 1; i <= n; i++){
        pp[i] = arr[i] - x;
    }   
    memset(dp, 0, sizeof(dp));
    memset(dp2, 0, sizeof(dp2));
    double num = -1.0*inf;
    for(int i = 1; i <= n; i++){
        dp[i] = max(dp[i], dp[i-1]+pp[i]);
        dp2[i] = min(dp2[i], dp2[i-1]+pp[i]);
        double f = abc(dp[i]), f2 = abc(dp2[i]);
        num = max(f, num);
        num = max(f2, num); 
    }   
    return num;
}

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    
    cin >> n;
    for(int i = 1; i <= n; i++){
        scanf("%lf", &arr[i]);
    }
    double l = -20000, r = 20000;
        
    for(int i = 1; i <= 100; i++){
        double lm = l + (r-l)/3;
        double rm = r - (r-l)/3;
        double lf = fun(lm);
        double rf = fun(rm);
        if (lf > rf) {
            l = lm;
            ans = min(ans, rf);
        } 
        else {
            r = rm;
            ans = min(ans, lf);
        }  
    }
    printf("%.15lf
", ans);    
    return 0;
}