对应位置的数字之间的关系是不变的，直接利用这一点就可以做
开始还写了ida*，结果t飞了

#include<map>
#include<queue>
#include<time.h>
#include<limits.h>
#include<cmath>
#include<ostream>
#include<iterator>
#include<set>
#include<stack>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep_1(i,m,n) for(int i=m;i<=n;i++)
#define mem(st) memset(st,0,sizeof st)
int read()
{
    int res=0,ch,flag=0;
    if((ch=getchar())=='-')             //判断正负
        flag=1;
    else if(ch>='0'&&ch<='9')           //得到完整的数
        res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9')
        res=res*10+ch-'0';
    return flag?-res:res;
}
typedef long long ll;
typedef pair<int,int> pii;
typedef unsigned long long ull;
typedef pair<double,double> pdd;
const int inf = 0x3f3f3f3f;
const int N=550;
int b[N];
int b1[N];
int a[N];
int a1[N];
map<pii,int>mp;
void solve()
{
    int n;
    cin>>n;
    mp.clear();
    for(int i=1; i<=n; i++)
        cin>>a[i],a1[i]=a[i];
    for(int i=1; i<=n; i++)
        cin>>b[i],b1[i]=b[i];
    sort(a1+1,a1+1+n);
    sort(b1+1,b1+1+n);
    for(int i=1; i<=n; i++)
        if(a1[i]!=b1[i])
        {
            cout<<"No"<<endl;
            return ;
        }
    for(int i=1; i<=(n+1)/2; i++)
        mp[ {min(a[i],a[n-i+1]),max(a[i],a[n-i+1]) } ] ++;
    for(int i=1; i<=(n+1)/2; i++)
        if(mp[ {min(b[i],b[n-i+1]),max(b[i],b[n-i+1]) } ]==0)
        {
            cout<<"No"<<endl;
            return ;
        }
        else
        {
            mp[ {min(b[i],b[n-i+1]),max(b[i],b[n-i+1]) } ] --;
            continue;
        }
    cout<<"Yes"<<endl;
    return ;

}
signed main()
{
    int t;
    cin>>t;
    while(t--)
        solve();
    return 0;
}