p03566 AtCoder Beginner Contest 076 - AtCoder Express
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
```python
n = int(input())
t = list(map(int, input().split()))
v = list(map(int, input().split()))
tmp = 0
e_t = []
for i in t:
    tmp+=i
    e_t.append(tmp)

def calc_end_speed(s):
    end_time = e_t[s]
    ma = 10**10
    for i in range(s+1, n):
        ma = min(e_t[i-1] - end_time + v[i], ma)
    return min(e_t[-1] - end_time, v[s], ma)

ans = 0
start_speed = 0

for sect in range(n):
    end_speed = min(calc_end_speed(sect), start_speed+t[sect])
    turn_time = (t[sect]+end_speed-start_speed)/2
    acce_end = min(v[sect]-start_speed, turn_time)
    dece_start = max(t[sect]-(v[sect]-end_speed) ,turn_time)
    max_speed = start_speed+acce_end
    ans += (start_speed+max_speed)*acce_end/2
    ans += max_speed*(dece_start-acce_end)
    ans += (end_speed+max_speed)*(t[sect]-dece_start)/2
    start_speed = end_speed

print(ans)
```