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 t[200];
double v[200];
double maxv[20000*2 + 1];

int main() {

    int n;

    cin >> n;

    t[0] = 0;

    for(int i=1; i<=n; i++){
        cin >> t[i];
        t[i] *= 2;
    }

    for(int i=1; i<=n; i++){
        t[i] += t[i-1];
    }

    for(int i=0; i<n; i++){
        cin >> v[i];
    }

    int end = t[n];

    for(int i=0; i<n; i++){
        for(int j=t[i]+1; j<=(t[i+1]); j++){
            maxv[j] = v[i];
        }
    }

    for(int i=0; i<n-1 ;i++){
        maxv[t[i+1]] = min(v[i],v[i+1]);
    }

    maxv[0] = 0;
    maxv[end] = 0;


    for(int i=1; i<=end; i++){
        maxv[i] = min(maxv[i],maxv[i-1] + 0.5);
    }

    for(int i=end-1; i>=0; i--){
        maxv[i] = min(maxv[i],maxv[i+1] + 0.5);
    }

    double ans = 0.0;

    for(int i=1; i<=end; i++){
        ans += (maxv[i] + maxv[i-1]) * 0.5 / 2.0;
    }

    cout << fixed << setprecision(9) << ans << endl;

    return 0;
}