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())
inpt=list(map(int,input().split()))
inpv=list(map(int,input().split()))

t=inpt+[0]
v=inpv+[0]

T=sum(t)
M=[0]

nowt=2*t[0]
nowmaxv=v[0]
nowv=M[0]
i=0

for time in range(2*T):
    if nowv+0.5<=v[i]:
        M.append(nowv+0.5)
    else:
        M.append(nowv)
    nowt-=1
    nowv=M[-1]
    if nowt==0:
        i+=1
        nowt=2*t[i]
        nowmaxv=v[i]
        nowv=min(nowv,nowmaxv)
        
t=list(reversed(inpt))+[0]
v=list(reversed(inpv))+[0]
m=[0]

nowt=2*t[0]
nowmaxv=v[0]
nowv=M[0]
i=0

for time in range(2*T):
    if nowv+0.5<=v[i]:
        m.append(nowv+0.5)
    else:
        m.append(nowv)
    nowt-=1
    nowv=m[-1]
    if nowt==0:
        i+=1
        nowt=2*t[i]
        nowmaxv=v[i]
        nowv=min(nowv,nowmaxv)

ans=[min(M[i],m[-i-1]) for i in range(len(M))]
print(sum([min(ans[i],ans[i+1])+(ans[i+1]-ans[i])**2 for i in range(len(ans)-1)])/2)