obtain array b from array a by sorting a range any number of

/* obtain array b from array a by sorting a range any number of times in a */

/* the trick will be to use sort opportunity as to sort two adjacent element so that a number comes to pos a from pos b without changing relative order of other elements */
void update(int node,int st,int en,int pos,int v)
{
    if(st>pos or en<pos)return ;
    if(st==en)
    {
        tree[node]=v;
        return ;
    }
    int mid=(st+en)>>1;
    update(node*2,st,mid,pos,v);
     update(node*2+1,mid+1,en,pos,v);
    tree[node]=min(tree[node*2],tree[node*2+1]);
}
int query(int node,int st,int en,int l,int r)
{
       if(st>r or en<l)return INT_MAX;
       if(st>=l and en<=r)return tree[node];
       int mid=(st+en)>>1;
       int v=query(node*2,st,mid,l,r);
       int w=query(node*2+1,mid+1,en,l,r);
       return min(v,w);
}
vector<int>s[mx];
int main()
{
    int i,j,k,l,m,n;
    scanf("%d",&m);
    while(m--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)s[i].clear();
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&ara[i]);
            update(1,1,n,i,ara[i]);
            s[ara[i]].pb(i);
        }
        for(int i=1;i<=n;i++)sort(s[i].rbegin(),s[i].rend());
        for(int i=1;i<=n;i++)
        {
           scanf("%d",&bara[i]);
        }
        int ck=1;
        for(int i=1;i<=n;i++)
        {

            int l=bara[i];
            if(s[l].size()==0)
            {
                ck=0;
                break;
            }
            int p=s[l][s[l].size()-1];
            s[l].pop_back();
            int k=query(1,1,n,1,p-1);
            if(k<l)ck=0;
            update(1,1,n,p,INT_MAX);
        }
        if(ck)printf("YES\n");
        else printf("NO\n");
    }

}