A - TL

Description

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, aseconds, 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, ..., a_n(1 ≤ a_i ≤ 100) — the running time of each of the n correct solutions in seconds. The third line contains m space-separated positive integers b₁, b₂, ..., b_m (1 ≤ b_i ≤ 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

Input

3 6
4 5 2
8 9 6 10 7 11

Output

5

Input

3 1
3 4 5
6

Output

-1

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题意：

找一个数满足大于第一行所有的数，小于第二行所有的数，至少大于等于第一行中一个数的2倍，找到符合条件的最小数

分析：

只要大于等于第一行最小数的2倍并且小于第二行最小值大于第一行最大值就行

代码：

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int a[105],b[105],n,m,i,k,l,t;
    while(cin>>n>>m)
    {
        for(i=0;i<n;i++)
            cin>>a[i];
        sort(a,a+n);
        k=2*a[0];
        t=a[n-1];
        for(i=0;i<m;i++)
            cin>>b[i];
        sort(b,b+m);
        l=b[0];
        //cout<<k<<endl<<t<<endl<<l<<endl;
        if(k<=t&&t<l)
            cout<<t<<endl;
        else
            if(k>t&&k<l)
            cout<<k<<endl;
        else
            cout<<"-1"<<endl;
    }
}

感受：

感觉这个题挺简单的，但是漏下一个地方，找不到错。。。冤死了，原因竟然还是因为懒，觉得写和不写没有差别。。。