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 ai (0 ≤ ai ≤ n - 1) — the statue currently placed on the i-th island. If ai = 0, then the island has no statue. It is guaranteed that the ai are distinct.

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

Output

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

Examples
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
Note

In the first sample, the islanders can first move statue 1 from island 1 to island 2, then move statue 2 from island 3 to island 1, and finally move statue 1 from island 2 to island 3.

In the second sample, the islanders can simply move statue 1 from island 1 to island 2.

In the third sample, no sequence of movements results in the desired position.

【题意】给你两个长度为n的数列，在这两个数列中有一个值为0的数，这些数可以通过0来改变自己现在的位置，问你能不能经过一些步骤，最后让a数列和b数列变的一样

【解题思路】开始还以为是逆序数的个数，果断归并一发，无情wa，然后自己各种找规律，实在找不到看了一发大牛的代码，恍然大悟，这题我做过呀。记得有一个求数组循环节的题和这个好类似，可当时就是没想出来，orz~太弱了，上分路漫漫，菜鸟要更加努力了。

【AC代码】

#include <bits/stdc++.h>
using namespace std;
const int nn = 200010;
int a[nn],b[nn],n,x;

int main()
{
    ios::sync_with_stdio();
    cin.tie(0);
    cin>>n;
    int cnt = 0,pos;
    for(int i=0; i<n; i++)
    {
        cin>>x;
        if(x==0)continue;
        a[cnt++] = x;
    }
    cnt = 0;
    for(int i=0; i<n; i++)
    {
        cin>>x;
        if(x==0)continue;
        b[cnt++] = x;
        if(x==a[0])pos = cnt-1;
    }
    bool flag=true;
    for(int i=0; i<n-1; i++)
    {
        if(a[i]!=b[(i+pos)%(n-1)])flag=false;
    }
    if(flag)puts("YES");
    else puts("NO");
    return 0;
}