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(lambda x: int(x), input().split()))
V = list(map(lambda x: int(x), input().split()))

dp = [[-float('inf')]*(101) for _ in range(N+1)]
V.append(0)

dp[0][0] = 1

for i in range(N):
    for v in range(V[i]+1):
        for vv in range(V[i]+1):
            past_dist = dp[i][v]
            if 2*V[i]-v-vv <= T[i]:
                a = V[i]-v
                b = V[i]-vv
                c = (v+V[i])*a/2 + V[i]*(T[i] - a - b) + (vv+V[i])*b/2
                dp[i+1][vv] = max(dp[i+1][vv], c+past_dist)
            elif T[i] >= abs(vv-v):
                ta = (vv-v+T[i])/2
                tb = (T[i]+v-vv)/2
                c = (2*v+ta)*ta/2 + (vv+v+ta)*tb/2
                dp[i+1][vv] = max(dp[i+1][vv], c+past_dist)

print(dp[N][0]-1)