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 PYTHON3):

Solution:
n=int(input())
t=list(map(int,input().split()))
v=list(map(int,input().split()))+[0]
for i in range(n-1,-1,-1):
    v[i]=min(v[i],v[i+1]+t[i])
ans=0
v0=0
for i in range(n):
    if v[i]<=v[i+1]:
        if v[i]-v0>t[i]:
            ans+=(2*v0+t[i])*t[i]/2
            v0=v0+t[i]
        else:
            ans+=t[i]*v[i]-(v[i]-v0)**2/2
            v0=v[i]
    else:
        if v[i]*2-v0-v[i+1]<t[i]:
            ans+=t[i]*v[i]-(v[i]-v0)**2/2-(v[i]-v[i+1])**2/2
            v0=v[i+1]
        elif v[i+1]-v0>t[i]:
            ans+=(2*v0+t[i])*t[i]/2
            v0=v0+t[i]
        else:
            d=abs(v0-v[i+1])
            ans+=max(v0,v[i+1])*t[i]-d**2/2+(t[i]-d)**2/4
            v0=v[i+1]
print(ans)