CodeForces 634A

Island Puzzle

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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Description

A remote island chain contains n islands, labeled 1 through n. Bidirectional bridges connect the islands to form a simple cycle — a bridge connects islands 1 and 2, islands 2 and 3, and so on, and additionally a bridge connects islands n and 1. The center of each island contains an identical pedestal, and all but one of the islands has a fragile, uniquely colored statue currently held on the pedestal. The remaining island holds only an empty pedestal.

The islanders want to rearrange the statues in a new order. To do this, they repeat the following process: First, they choose an island directly adjacent to the island containing an empty pedestal. Then, they painstakingly carry the statue on this island across the adjoining bridge and place it on the empty pedestal.

Determine if it is possible for the islanders to arrange the statues in the desired order.

Input

The first line contains a single integer n (2 ≤ n ≤ 200 000) — the total number of islands.

The second line contains n space-separated integers a_i (0 ≤ a_i ≤ n - 1) — the statue currently placed on the i-th island. If a_i = 0, then the island has no statue. It is guaranteed that the a_i are distinct.

The third line contains n space-separated integers b_i (0 ≤ b_i ≤ n - 1) — the desired statues of the ith island. Once again, b_i = 0indicates the island desires no statue. It is guaranteed that the b_i are distinct.

Output

Print "YES" (without quotes) if the rearrangement can be done in the existing network, and "NO" otherwise.

Sample Input

Input

3
1 0 2
2 0 1

Output

YES

Input

2
1 0
0 1

Output

YES

Input

4
1 2 3 0
0 3 2 1

Output

NO

解体思路：我们可以把0号岛屿忽略， 因为0的存在位置不会改变其他岛屿的相对位置。那么只要判断最终的岛屿序列（不含0），是否是原序列或原循环序列（例如2，3，1 2，3，1，2，3，1）的子序列就行了（不含0）。

代码如下：

#include<stdio.h>
int x1[200010];//第i号岛屿含有某种类型的雕塑 
int x2[200010];
int main(){
	int n,k,i,j;
	int flag,p; 
	while(scanf("%d",&n)!=EOF){
		k=0;
		for(i=0;i<n;i++){
			scanf("%d",&x1[k]);
			if(x1[k]==0)continue;
			else k++;
		}
		k=0;
		for(i=0;i<n;i++){
			scanf("%d",&x2[k]);
			if(x2[k]==0)continue;
			else k++;
	   }
	   for(i=0;i<k;i++){
	   	 if(x1[i]==x2[0]){
	   	 	j=i;
	   	 	break;
	   	 }
	   }
	   p=0;flag=0;
		while(1){
			if(j==k){
				j=0;
			}
			if(p==k){
				flag=1;
				break;
			}
			if(x1[j]==x2[p]){
				p++;
				j++;
			}
			else break;
		}
		if(flag)printf("YES\n");
		else printf("NO\n");
	}
	return 0;
}