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())
*t, = map(int, input().split())
*v, = map(int, input().split())
sumt = sum(t)

sv = [0]
for i in range(N):
    for j in range(t[i]*2):
        sv.append(v[i])
sv.append(0)

# dp[i][j] := i/2秒後に速度がj/2のときの最大距離
dp = [[-float('inf')]*201 for _ in range(sumt*2+1)]
dp[0][0] = 0
for i in range(1, len(dp)):
    maxj = min(sv[i], sv[i+1])*2
    for j in range(len(dp[i])):
        k = 0.25*j
        # 変化なし
        if j <= maxj:
            dp[i][j] = max(dp[i][j], dp[i-1][j]+k)
        # 加速
        if (j+1) <= maxj:
            dp[i][j+1] = max(dp[i][j+1], dp[i-1][j]+k+0.25)
        # 減速
        if j > 0 and (j-1) <= maxj:
            dp[i][j-1] = max(dp[i][j-1], dp[i-1][j]+k-0.25)

print(dp[-1][0])
