Chirag wanted to prepare a Programmin round. He's already got one problem and he wants to set a time limit (TL) on it.

Chirag has written n correct solutions. For each correct solution, he knows its running time (in seconds). Chirag has also wrote m wrong solutions and for each wrong solution he knows its running time (in seconds).

Let's suppose that Chirag will set v seconds TL in the problem. Then we can say that a solution passes the system testing if its running time is at most v seconds. We can also say that a solution passes the system testing with some "extra" time if for its running time, a seconds, an inequality 2a ≤ v holds.

As a result, Chirag decided to set v seconds TL, that the following conditions are met:

v is a positive integer;
all correct solutions pass the system testing;
at least one correct solution passes the system testing with some "extra" time;
all wrong solutions do not pass the system testing;
value v is minimum among all TLs, for which points 1, 2, 3, 4 hold.

Help Chirag and find the most suitable TL or else state that such TL doesn't exist.
Input

The first line contains two integers n, m (1 ≤ n, m ≤ 100). The second line contains n space-separated positive integers a1, a2, ..., an (1 ≤ ai ≤ 100) — the running time of each of the n correct solutions in seconds. The third line contains m space-separated positive integers b1, b2, ..., bm (1 ≤ bi ≤ 100) — the running time of each of m wrong solutions in seconds.
Output

If there is a valid TL value, print it. Otherwise, print -1.

SAMPLE INPUT
3 6
4 5 2
8 9 6 10 7 11

SAMPLE OUTPUT
5

Difficulty: UNKNOWN_DIFFICULTY

Solutions:
'''
# Read input from stdin and provide input before running code
name = raw_input('What is your name?\n')
print 'Hi, %s.' % name
'''
#print 'Hello World!'
n = list(map(int,input().split()))
a = list(map(int,input().split()))
b = list(map(int,input().split()))
a.sort()
b.sort()
magic = max(2*a[0], a[len(a)-1]);
if magic<b[0]:
	print(magic)
else:
	print(-1)