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;


int main()
{
    int N;
    cin >> N;
    vector<double> t(N+2, 0), v(N+2, 0);
    for(int i = 1; i <= N; i++) cin >> t[i];
    for(int i = 1; i <= N; i++) cin >> v[i];
 
 
    vector<double> l(N+2, 0), r(N+2, 0);
    for(int i = 2; i < N+2; i++) l[i] = l[i-1] + t[i-1];
    for(int i = 1; i < N+2; i++) r[i] = r[i-1] + t[i];

    vector<double> maxv;
    for(double x = 0; x <= r[N+1]; x += 0.5) {
        double minv = 100;
        for(int i = 0; i < N+2; i++) {
            if( x <= l[i] )           minv = min( minv, v[i] + (l[i] - x) );
            if( l[i] < x && x < r[i]) minv = min( minv, v[i] );
            if( r[i] <= x )           minv = min( minv, v[i] + (x - r[i]) );
        }
        maxv.push_back( minv );
    }
 
    double ans = 0;
    for(int i = 1; i < maxv.size(); i++) ans += 0.5 * (maxv[i] + maxv[i-1]) / 2;
    cout << fixed << setprecision(3) << ans << endl;
 
    return 0;
}