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(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout.precision(12);
    cout.setf(ios_base::fixed, ios_base::floatfield);
    
    int n, t[100], v[101], st[101];
    cin >> n;
    for(int i=0;i<n;i++) cin >> t[i];
    for(int i=0;i<n;i++) cin >> v[i];
    v[n] = 0;

    st[0] = 0;
    for(int i=1;i<=n;i++) st[i] = st[i-1] + t[i-1];

    int vel[40001];
    int pos = 0;
    vel[0] = 0;
    for(int i=1;i<=st[n]*2;i++){
        vel[i] = min(vel[i-1]+1, v[pos]*2);
        for(int j=pos+1;j<=n;j++) vel[i] = min(vel[i], v[j]*2 + 2*st[j]-i);
        if(i == st[pos+1]*2) pos++;
    }

    double ans = 0;
    for(int i=1;i<=st[n]*2;i++){
        if(vel[i-1] != vel[i]) ans += 1;
        ans += min(vel[i-1], vel[i]) * 2;        
    }
    cout << ans/8 << endl;
    return 0;
}