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, os

f = lambda:list(map(int,input().split()))
if 'local' in os.environ :
    sys.stdin = open('./input.txt', 'r')

def solve():
    n = f()[0]
    t = f()
    v = f()
    


    sumt = sum(t)*2
    vs = [100000000] * (sumt+1)

    vs[0] = 0

    for i in range(1,sumt):
        totalt = 0
        idx = 0
        for k in range(n):
            totalt += t[k]*2
            if totalt >= i:
                idx = k
                break

        vs[i] = min(vs[i-1]+0.5, v[idx])

    vs[sumt] = 0
    for i in range(sumt-1, 0, -1):
        totalt = 0
        idx = 0
        for k in range(n):
            totalt += t[k]*2
            if totalt >= i+1:
                idx = k
                break

        vs[i] = min(vs[i+1]+0.5, vs[i], v[idx])

    ans = 0
    for i in range(sumt):
        ans += (vs[i]+ vs[i+1])/4


    # print(dict(zip(range(len(vs)), vs)))
    print(ans)

solve()
