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()]
v = [int(i) for i in input().split()]

maxspeed = [0]*(n+1)
for i in range(n-1,0,-1):
    maxspeed[i] = min(v[i], v[i-1], t[i] + maxspeed[i+1])

def dist(i,first): # distance,lastspeed
    ta = v[i] - first
    tb = v[i] - maxspeed[i+1]
    if ta + tb <= t[i]:
        c = (first+v[i])*ta/2 + v[i]*(t[i]-ta-tb) + (maxspeed[i+1]+v[i])*tb/2
        lastspeed = maxspeed[i+1]
    elif maxspeed[i+1] - first > t[i]:
        c = (first + first + t[i])*t[i]/2
        lastspeed = first + t[i]
    else:
        a = (maxspeed[i+1]+t[i]-first)/2
        c = (first+first+a)*a/2 + (first+a+maxspeed[i+1])*(t[i]-a)/2
        lastspeed = maxspeed[i+1]
    return c, lastspeed


ans = 0
firstspeed = 0
for i in range(n):
    d,firstspeed = dist(i,firstspeed)
    ans += d

print(ans)