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 = [int(i) for i in input().split()]
L = [0]
for i in range(N):
    L.append(L[i] + T[i])
V = [int(i) for i in input().split()]

def VM(x, i): #x秒後に条件iがとり得る速度
    if 0 <= x <= L[i]:
        return V[i] + (L[i] - x)
    elif L[i] < x <= L[i+1]:
        return V[i]
    else:
        return V[i] + (x - L[i+1])

Search = [[] for i in range(N)] #設定された最高速度よりも小さくなり得る条件を探索
for i in range(N):
    for j in range(N):
        if i < j and V[i] - V[j] > L[j] - L[i+1]:
            Search[i].append(j)
        elif i > j and V[i] - V[j] > L[i] - L[j+1]:
            Search[i].append(j)

maxV = [float("inf") for i in range(2*L[-1]+1)]
for i in range(N):
    for j in range(2 * L[i], 2 * L[i+1] + 1):
        maxV[j] = min(V[i], j * 0.5, L[-1] - j * 0.5) #初期化
        for s in Search[i]:
            maxV[j] = min(maxV[j], VM(j * 0.5, s))
Ans = 0
for i in range(2*L[-1]):
    Ans += (maxV[i] + maxV[i+1])/4
print(Ans)