codeforces A. TL 题解

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

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

Let's suppose that Valera 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, Valera decided to set v seconds TL, that the following conditions are met:

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

Help Valera 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 mwrong solutions in seconds.

Output

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

Sample test(s)

input

3 64 5 28 9 6 10 7 11

output

5

这句话难读懂：

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.

就是说：如果一个AC的解决方案的运行时间是a，满足2*a <= v，那么就说这个运行时间通过而且有"extra"时间(a)

void TLjudge(){int n, m, minVal = 1<<30, maxVal = 1<<31, a = 0;cin>>n>>m;while (n--){cin>>a;if (a < minVal) minVal = a;if (a > maxVal) maxVal = a;}int ans = max(minVal*2, maxVal);bool ok = true;while (m--){cin>>a;if (a <= ans){ok = false;break;}}if (ok) cout<<ans;else cout<<-1;};