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;

//https://atcoder.jp/contests/abc076/submissions/1696628
int n, t[109], v[109],s[109],p[40009];

int main(){
    scanf("%d",&n);
    for (int i=1;i<=n;i++) {
        scanf("%d",&t[i]);
        t[i]*=2;
        s[i+1]=s[i]+t[i];
    }
    for (int i=1;i<=n;i++) {
        scanf("%d",&v[i]);
        v[i]*=2;
    }
    for(int i=0;i<=s[n+1];i++) {
        for (int j=1;j<=n+1;j++) {
            if (s[j]>=i) {
                p[i]=v[j-1];
                break;
            }
        }
        for (int j=0;j<=n+1;j++) {
            int dst=abs(s[j]-i);
                p[i]=min(p[i],dst+min(v[j-1],v[j]));
        }
    }
    int sum=0;
    for (int i=0;i<=s[n+1];i++) sum+=p[i];
    printf("%.12Lf\n",0.25l*sum);
    return 0;
}