题干：

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.

Examples

Input

4
2 1 3
2 4 2

Output

6 2

Input

3
1 2
2 1 3

Output

-1

Note

First sample:

Second sample:

 

题目大意：

    两个人玩纸牌游戏，每一轮游戏每个人都把自己的第一个牌拿出来比较大小，大的一方先将对面的这个牌放在自己牌的最后，再把自己的牌放在最后，直到一个人没有了纸牌就算那个人输，问一共会进行多少次游戏，以及赢的人的编号。如果决不出胜负那就输出-1

解题报告：

   模拟就行了。猜他最多不超过1e7次、、其实这题的上界是4e7次，因为你想啊，最多的排列可能是n!种，然后分到两个人手中，可能每个人手中的牌有(0,n)(1,n-1)...(n,0)共(n+1)中可能性，所以一共的局面有(n+1)!种情况，算一下最多4e7左右。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#include<ctime>
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
const int MAX = 2e5 + 5;
int n,k1,k2;
deque<int> a,b;
int main()
{
	cin>>n;
	cin>>k1;
	for(int x,i = 1; i<=k1; i++) {
		scanf("%d",&x);a.pb(x);
	}
	cin>>k2;
	for(int x,i = 1; i<=k2; i++) {
		scanf("%d",&x);b.pb(x);
	}
	if(k1 == 0) return 0 * printf("0 2\n");
	if(k2 == 0) return 0 * printf("0 1\n");
	int i;
	int flag = 0;
	for(i = 1; i<=10000000; i++) {
		int aa = a.front();a.pop_front();
		int bb = b.front();b.pop_front();
		if(aa > bb) {
			a.pb(bb);a.pb(aa);
		}
		else {
			b.pb(aa);b.pb(bb);
		}
		if(a.empty() || b.empty()) {
			flag = 1;break;
		}
		//printf("a : %d b : %d\n",a.size(),b.size());
	}
	if(a.empty()) {
		printf("%d %d\n",i,2);
	}
	if(b.empty()) {
		printf("%d %d\n",i,1);
	}
	if(i > 10000000) printf("-1\n");
	return 0 ;
 }