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
"""
dp[t][v] = 時刻tに速度vで走っている時の今までの走行距離の最大値
dp[0][0] = 0, dp[0][i] = -inf(i>0)
条件を考える
  dp[t][v]について、時刻tにおける速度の最大値をM(t)とすると、
  v >  M[t] -> dp[v][t] = -inf
  v <= M[t] -> dp[v][t] = max(dp[t-1][v+1]+(2*v+1)/2, dp[t-1][v]+v, dp[t-1][v-1]+(2*v-1)/2)
ans = max(dp[sum(T)])
"""
from itertools import accumulate
U = 300
n = int(input())
T = [0] + list(map(lambda x:int(x)*2, input().split()))
V = list(map(lambda x:int(x)*2, input().split())) + [0]
S = sum(T)
M = [0]*(S+1)
K = list(accumulate(T))
for i in range(n):
  for j in range(K[i]+1, K[i+1]):
    M[j] = V[i]
  M[K[i+1]] = min(V[i], V[i+1])
dp = [[-float("inf")]*(U+1) for _ in range(S+1)]
dp[0][0] = 0
for t in range(1, S+1):
  for v in range(U+1):
    if v > M[t]:
      continue
    dp[t][v] = dp[t-1][v] + v
    if v > 0:
      dp[t][v] = max(dp[t][v], dp[t-1][v-1]+(2*v-1)/2)
    if v < U:
      dp[t][v] = max(dp[t][v], dp[t-1][v+1]+(2*v+1)/2)
ans = max(dp[S]) / 4
print(ans)