codeforces B. Island Puzzle

B. Island Puzzle

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 = 0indicates 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.

思路：KMP

其实0可以省略看成没有，去0后不改变原来的顺序，其实只要适当移动0可以移到任意位置，所以把0全移到最前这样就能看成没有02020---00022。然后把去0后的原窜复制一遍放在后面，然后下面的窜去0后对上串KMP即可。O（n）；

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
#include<math.h>
#include<queue>
#include<map>
using namespace std;
typedef long long LL;
void next(int n);
int aa[200005];
int a[400005];
int bb[200005];
int f[200005];
int nex[200005];
int main(void)
{
    int i,j,k,p,q;
    cin>>k;int cnt=0;
    for(i=0;i<k;i++)
    {
        cin>>p;
        if(p!=0)
        {cnt++;a[cnt]=aa[cnt]=p;}
    }
    for(i=cnt+1;i<=2*cnt;i++)
    {
        a[i]=aa[i-cnt];
    }
    int ans=0;
    for(i=0;i<k;i++)
    {
        cin>>p;
        if(p!=0)
        { ans++;
            bb[ans]=p;

        }
    }
    next(ans);
    j=0;int flag=0;
    for(i=1;i<=2*cnt;i++)
    {
        while(j>0&&bb[j+1]!=a[i])
        {
            j=nex[j];
        }
        if(bb[j+1]==a[i])
        {
            j++;
        }
        if(j==ans)
        {flag=1;
            break;
        }
    }if(flag)
    {
        printf("YES
");
    }
    else printf("NO
");
    return 0;
}
void next(int n)
{
    int i,j;
    nex[0]=0;
    nex[1]=0;j=0;
    for(i=2;i<=n;i++)
    {
        while(bb[j+1]!=bb[i]&&j>0)
        {
            j=nex[j];
        }
        if(bb[j+1]==bb[i])
        {
            j++;
        }
        nex[i]=j;
    }
}