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>
typedef long long ll;
using namespace std;
int INF = 1LL << 30;
int MOD = 1e9+7;
int main(){
    int N;
    cin >> N;
    vector<int> L(N+2),R(N+2),V(N+2);
    int T = 0;
    //l(s) - r(s)
    for(int i = 1;i <= N;i++){
        int t;
        cin >> t;
        t*=2;//
        L[i] = R[i-1];       
        R[i] = L[i] + t;
        T += t;
    }
    for(int i = 1;i <= N;i++)cin >> V[i],V[i]*=2;
    L[N+1] = R[N+1] = T;    
    vector<double> vel;
    for(int i = 0;i <= T;i++){
        double v = INF;
        int t = i;
        for(int j = 0;j < N+2;j++){
            if      (t <= L[j])   v = min(v, 1.*V[j] + L[j] - t);
            else if (t >= R[j])   v = min(v, 1.*V[j] + t - R[j]);
            else                  v = min(v, 1.*V[j]);
        }
        vel.push_back(v);
    }
    double sum = 0;
    for(int i = 0;i < T;i++){sum += 0.25*(vel[i] + vel[i+1]);}
    printf("%.9lf\n",sum/2);
}