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 <iostream>
#include <algorithm>
using namespace std;
 
int main()
{
    int n;
    cin >> n;
    int t[102], v[102];
    int l[102], r[102];
    int pt[102];
    int d[40003];
    l[0] = 0;
    pt[0] = 0;
    for(int i = 1; i <= n; i++){
        cin >> t[i];
        t[i] *= 2;
        pt[i] = pt[i - 1] + t[i];
    }
    for(int i = 1; i <= n; i++){
        cin >> v[i];
        v[i] *= 2;
        l[i] = min(l[i - 1], v[i] - pt[i]);
    }
    r[n] = 0;
    for(int i = n - 1; i >= 0; i--){
        r[i] = min(r[i + 1], v[i + 1] - (pt[n] - pt[i]));
    }
    for(int i = 1; i <= n; i++){
        for(int j = pt[i - 1]; j <= pt[i]; j++){
            d[j] = min(l[i - 1] + j, r[i] + (pt[n] - j));
            d[j] = min(d[j], v[i]);
        }
    }
    double ans = 0;
    for(int i = 0; i < pt[n]; i++){
        ans += d[i] + d[i + 1];
    }
	printf("%.15f\n", ans / 8.0);
}