Problem: 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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
using ll = long long;

int main()
{
    int N;
    cin >> N;
    double L[N+2], R[N+2], V[N+2];
    L[0] = R[0] = V[0] = 0;
    double tm = 0;
    for(int i = 1; i <= N; ++i) {
        double t; cin >> t;
        L[i] = R[i-1];
        R[i] = L[i] + t;
        tm += t;
    }
    L[N+1] = R[N+1] = tm;
    V[N+1] = 0;
    for(int i = 1; i <= N; ++i) {
        double v; cin >> v;
        V[i] = v;
    }

    double res = 0;
    double v0 = 0, v1 = 400;
    for(int i = 1; (double)i*0.5 <= tm; ++i) {
        double x = (double)i*0.5;
        for(int j = 0; j < N+2; ++j) {
            double l = L[j], r = R[j], v = V[j];
            if(x <= l) v1 = min(v1, v + (l - x));
            else if(l <= x && x <= r) v1 = min(v1, v);
            else v1 = min(v1, v + (x - r));
        }
        res += 0.25*(v0 + v1);
        v0 = v1, v1 = 400;
    }
    cout << fixed << setprecision(10) << res << endl;
    return  0;
}