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 = [0] + [int(x) for x in input().split()]
v = [0] + [int(x) for x in input().split()]

max_speed_from_left = [0] * (N+1) # index i = t[i],t[i+1]の間
max_speed_from_right = [0] * (N+1) # index i = t[i],t[i+1]の間

for i in range(1,N):
  start = max_speed_from_left[i-1]
  max_speed_from_left[i] = min(start + t[i], v[i], v[i+1])
  
for i in range(N-1,0,-1):
  start = max_speed_from_right[i+1]
  max_speed_from_right[i] = min(start + t[i+1], v[i], v[i+1])

speed = [min(x,y) for x,y in zip(max_speed_from_left,max_speed_from_right)]

def dist(left_speed,right_speed,t,v):
  x = (v - left_speed) + (v - right_speed)
  if x >= t:
    v = (left_speed + right_speed + t)/2
    x = t
  # 最高速度vに到達して、t-xあまる
  d = (left_speed + v) * (v-left_speed)/2
  d += v*(t-x)
  d += (right_speed + v) * (v-right_speed)/2
  return d

answer = 0
for i in range(1,N+1):
  answer += dist(speed[i-1],speed[i],t[i],v[i])
print(answer)