队列 Soldier and Cards

Description

Two bored soldiers are playing card war. Their card deck consists of exactly n cards, numbered from 1 to n, all values are different. They divide cards between them in some manner, it's possible that they have different number of cards. Then they play a "war"-like card game.

The rules are following. On each turn a fight happens. Each of them picks card from the top of his stack and puts on the table. The one whose card value is bigger wins this fight and takes both cards from the table to the bottom of his stack. More precisely, he first takes his opponent's card and puts to the bottom of his stack, and then he puts his card to the bottom of his stack. If after some turn one of the player's stack becomes empty, he loses and the other one wins.

You have to calculate how many fights will happen and who will win the game, or state that game won't end.

Input

First line contains a single integer n (2 ≤ n ≤ 10), the number of cards.

Second line contains integer k₁ (1 ≤ k₁ ≤ n - 1), the number of the first soldier's cards. Then follow k₁ integers that are the values on the first soldier's cards, from top to bottom of his stack.

Third line contains integer k₂ (k₁ + k₂ = n), the number of the second soldier's cards. Then follow k₂ integers that are the values on the second soldier's cards, from top to bottom of his stack.

All card values are different.

Output

If somebody wins in this game, print 2 integers where the first one stands for the number of fights before end of game and the second one is 1 or 2 showing which player has won.

If the game won't end and will continue forever output - 1.

Sample Input

Input

4             输入总共有几张
2 1 3         先输入第一个人有几张牌，再输入这几张牌的牌值
2 4 2         先输入第二个人有几张牌，再输入这几张牌的牌值

Output

6 2           输出比较的次数和胜利的人

Input

3
1 2
2 1 3

Output

-1

Hint

First sample:

$\text{[math]}$

Second sample: $\text{[math]}$

题意：

两个人每个人都有一堆牌，他们每个人从他那堆拿出最上面的牌,并放在桌子上。

牌值更大的那个先他的对手的牌他的牌的底部,然后他把他的卡片放他牌的底部，

如此循环。如果有一个玩家的一个堆栈为空,他输了,另一个胜利。

思路：

将两个人的牌值分别放入不同的队列，后队首与队首进行比较，将值小的那队的对首放入，另一队的队尾再将值大的那个队的队首放队尾。

将两个队的队首都删除。如此循环直到其中一个队为空时跳出循环。

知识：

队列提供了下面的操作

q.empty() 如果队列为空返回true，否则返回false
q.size() 返回队列中元素的个数
q.pop() 删除队列首元素但不返回其值
q.front() 返回队首元素的值，但不删除该元素
q.push() 在队尾压入新元素
q.back() 返回队列尾元素的值，但不删除该元素

 1 #include<iostream>
 2 #include<queue>
 3 #include<cstring>
 4 using namespace std;
 5 int main()
 6 {
 7    int n,k1,k2,i,k,b;
 8    queue<int>c,d;
 9    cin>>n;
10    cin>>k1;
11 for(i=0;i<k1;i++)
12 {    cin>>b;
13    c.push(b);}
14    cin>>k2;
15 for(i=0;i<k2;i++)
16 {cin>>b;
17 d.push(b);}
18 k=0;
19 while(1)
20 {if(c.empty()||d.empty()) break;
21 
22 if(c.front()>d.front())
23 {c.push(d.front());
24  c.push(c.front());
25  d.pop();
26  c.pop();
27 }
28 else 
29 {d.push(c.front());
30  d.push(d.front());
31  d.pop();
32  c.pop();
33 }
34     k++;
35 if(k>1e6)
36     {    cout<<"-1"<<endl;
37     break;}
38 
39 }
40 if(c.empty())
41   cout<<k<<" 2"<<endl;
42 else  if(k<=1e6)  
43 cout<<k<<" 1"<<endl;
44 return 0;
45 }

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