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
n = int(input())
ts = [int(c) for c in input().split()]
vs = [int(c) for c in input().split()]

t = sum(ts)
dp = [[-10 ** 10 for _ in range(101)] for _ in range(t + 1)]
dp[0][0] = 0.

mvs = []
for dt, mv in zip(ts, vs):
    for st in range(dt):
        mvs.append(mv)
mvs.append(0)

for ct in range(t):
    mv = mvs[ct]
    for cv in range(mv + 1):
        endmv = min(mv, mvs[ct+1])
        for dv in [-1, 0, 1]:
            nv = cv + dv
            if 0 <= nv <= endmv:
                dp[ct + 1][nv] = max(dp[ct][cv] + (cv + nv) / 2, dp[ct + 1][nv])
        if mv >= cv + 1 and cv <= endmv:
            dp[ct + 1][cv] = max(dp[ct+1][cv], dp[ct][cv] + cv + 0.25)


print(dp[t][0])
