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
import sys

sys.setrecursionlimit(10 ** 6)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]

def main():
    n=int(input())
    tt=LI()
    vv=LI()+[0]
    #区間の変わり目においてのvの最大値を左右から調べる
    y=0
    max_vL=[1000]*(n-1)
    max_vR=[1000]*(n-1)
    for i in range(n-1):
        y=min(vv[i],vv[i+1],y+tt[i])
        max_vL[i]=y
    y=0
    for i in range(n-2,-1,-1):
        y=min(vv[i+1],vv[i],y+tt[i+1])
        max_vR[i]=y
    #print(max_vR)
    #print(max_vL)
    #左右のminを取ることで、変わり目のvを決める
    yy=[0]*(n+1)
    for i in range(n-1):
        yy[i+1]=min(max_vL[i],max_vR[i])
    #print(yy)
    #各区間の面積を求める
    ans=0
    for t,y0,y1,v in zip(tt,yy,yy[1:],vv):
        h=(t+y0+y1)/2
        ans+=((h+y0)*(h-y0)+(h+y1)*(h-y1))/2
        if h>v:ans-=(h-v)**2
    print(ans)

main()