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;
#define rep(i,n) for(int i=0;i<int(n);++i)

int main(void){
    int N;
    cin>>N;
    vector<int> t(N);
    vector<double> v(N);
    int T=0;
    rep(i,N) {
        cin>>t[i];
        T += t[i]*2;
    }
    rep(i,N) cin>>v[i];
    vector<double> s(T+1);
    rep(x,T){
        if(x<=T/2) s[x] = x/2.0;
        else s[x] = (T-x)/2.0;
    }
    // rep(x,T) cout<<s[x]<<endl;
    int l=0,r=0;
    rep(i,N){
        r += 2*t[i];
        rep(x,T){
            if(x<=l) s[x] = min(s[x],v[i]+(l-x)/2.0);
            else if(x<=r) s[x] = min(s[x],v[i]);
            else s[x] = min(s[x],v[i]+(x-r)/2.0);
        }
        l = r;
    }

    double sum = 0;
    rep(x,T){
        sum += 0.25*(s[x]+s[x+1]);
    }
    printf("%.6f\n",sum);

    return 0;
}