題干：

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:

?

題目大意：

? ??兩個人玩紙牌游戲，每一輪游戲每個人都把自己的第一個牌拿出來比較大小，大的一方先將對面的這個牌放在自己牌的最后，再把自己的牌放在最后，直到一個人沒有了紙牌就算那個人輸，問一共會進行多少次游戲，以及贏的人的編號。如果決不出勝負那就輸出-1

解題報告：

? ?模擬就行了。猜他最多不超過1e7次、、其實這題的上界是4e7次，因為你想啊，最多的排列可能是n!種，然后分到兩個人手中，可能每個人手中的牌有(0,n)(1,n-1)...(n,0)共(n+1)中可能性，所以一共的局面有(n+1)!種情況，算一下最多4e7左右。

AC代碼：

#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 ;}

?

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