In the year 2168, AtCoder Inc., which is much larger than now, is starting a limited express train service called AtCoder Express.

In the plan developed by the president Takahashi, the trains will run as follows:

* A train will run for (t_1 + t_2 + t_3 + ... + t_N) seconds.
* In the first t_1 seconds, a train must run at a speed of at most v_1 m/s (meters per second). Similarly, in the subsequent t_2 seconds, a train must run at a speed of at most v_2 m/s, and so on.



According to the specifications of the trains, the acceleration of a train must be always within ±1m/s^2. Additionally, a train must stop at the beginning and the end of the run.

Find the maximum possible distance that a train can cover in the run.

Constraints

* 1 \leq N \leq 100
* 1 \leq t_i \leq 200
* 1 \leq v_i \leq 100
* All input values are integers.

Input

Input is given from Standard Input in the following format:


N
t_1 t_2 t_3 … t_N
v_1 v_2 v_3 … v_N


Output

Print the maximum possible that a train can cover in the run.
Output is considered correct if its absolute difference from the judge's output is at most 10^{-3}.

Examples

Input

1
100
30


Output

2100.000000000000000


Input

2
60 50
34 38


Output

2632.000000000000000


Input

3
12 14 2
6 2 7


Output

76.000000000000000


Input

1
9
10


Output

20.250000000000000000


Input

10
64 55 27 35 76 119 7 18 49 100
29 19 31 39 27 48 41 87 55 70


Output

20291.000000000000
#include <bits/stdc++.h>
using namespace std;

int main(){
    int N;
    cin >> N;
    vector<double> t(N+2, 0);
    vector<double> v(N+2, 0);
    for (int i = 1; i <= N; i++){
        double n;
        cin >> n;
        t[i] = t[i-1] + 2*n;
    }
    t[N+1] = t[N];
    for (int i = 1; i <= N; i++){
        cin >> v[i];
    }
    
    vector<double> s(t[N], 100000);
    for (int i = 0; i <= t[N]; i++) {
        for (int j = 0; j <= N+1; j++) {
            double f;
            if (i >= t[j]) {
                f = v[j] + 0.5 * (i-t[j]);
            }else if (i >= t[j-1]) {
                f = v[j];
            }else {
                f = v[j] + 0.5 * (t[j-1] - i);
            }
            s[i] = min(s[i], f);
        }
    }
    double ans = 0;
    for (int i = 1; i <= t[N]; i++) {
        ans += 0.25 * (s[i] + s[i-1]);
    }
    printf("%.5lf", ans);
    return 0;
}