You are given two permutations p and q, consisting of n elements, and m queries of the form: l1, r1, l2, r2 (l1 ≤ r1; l2 ≤ r2). The response for the query is the number of such integers from 1 to n, that their position in the first permutation is in segment [l1, r1] (borders included), and position in the second permutation is in segment [l2, r2] (borders included too).

A permutation of n elements is the sequence of n distinct integers, each not less than 1 and not greater than n.

Position of number v (1 ≤ v ≤ n) in permutation g1, g2, ..., gn is such number i, that gi = v.

Input

The first line contains one integer n (1 ≤ n ≤ 106), the number of elements in both permutations. The following line contains n integers, separated with spaces: p1, p2, ..., pn (1 ≤ pi ≤ n). These are elements of the first permutation. The next line contains the second permutation q1, q2, ..., qn in same format.

The following line contains an integer m (1 ≤ m ≤ 2·105), that is the number of queries.

The following m lines contain descriptions of queries one in a line. The description of the i-th query consists of four integers: a, b, c, d (1 ≤ a, b, c, d ≤ n). Query parameters l1, r1, l2, r2 are obtained from the numbers a, b, c, d using the following algorithm: 

  1. Introduce variable x. If it is the first query, then the variable equals 0, else it equals the response for the previous query plus one. 
  2. Introduce function f(z) = ((z - 1 + x) mod n) + 1. 
  3. Suppose l1 = min(f(a), f(b)), r1 = max(f(a), f(b)), l2 = min(f(c), f(d)), r2 = max(f(c), f(d)). 

Output

Print a response for each query in a separate line.

Examples

Input

3
3 1 2
3 2 1
1
1 2 3 3


Output

1


Input

4
4 3 2 1
2 3 4 1
3
1 2 3 4
1 3 2 1
1 4 2 3


Output

1
1
2
1
1
1
1
1 1 1 1

1

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
using namespace std;
const int Maxn=1000005;
int n,m,pos[Maxn],ans=-1;

inline int get() {
    char ch;
    while (!isdigit(ch=getchar()));
    int v=ch-48;
    while (isdigit(ch=getchar())) v=v*10+ch-48;
    return v;
}

struct node{
    int sum;
    node *lc,*rc;
}po[Maxn*21],*pn=po,*null=po;

class Segment_Tree{
    node *root;
    void ins(node *&x,node *y,int l,int r,int p) {
        (x=++pn)->sum=y->sum+1;
        if (l==r) return;
        int mid=(l+r)>>1;
        if (p<=mid) x->rc=y->rc,ins(x->lc,y->lc,l,mid,p);
        else x->lc=y->lc,ins(x->rc,y->rc,mid+1,r,p);
    }
    int qry(node *x,node *y,int l,int r,int s,int t) {
        int mid,res=0;
        while (s>l||t<r) {
            mid=(l+r)>>1;
            if (t<=mid) x=x->lc,y=y->lc,r=mid;
            else if (s>mid) x=x->rc,y=y->rc,l=mid+1;
            else res=qry(x->lc,y->lc,l,mid,s,t),x=x->rc,y=y->rc,l=mid+1;
        }
        return res+x->sum-y->sum;
    }
    public:
        void set() {
            null->lc=null->rc=null;
            root=null;
        }
        void ins(Segment_Tree &pre,int p) {
            ins(root,pre.root,1,n,p);
        }
        int qry(Segment_Tree &pre,int s,int t) {
            return qry(root,pre.root,1,n,s,t);
        }
}tr[Maxn];

inline int f(int x) {
    return (x+ans)%n+1;
}

int main() {
    n=get();
    for (int i=1;i<=n;i++) pos[get()]=i;
    tr[0].set();
    for (int i=1;i<=n;i++) tr[i].ins(tr[i-1],pos[get()]);
    m=get();
    for (int i=1;i<=m;i++) {
        int a=f(get()),b=f(get()),c=f(get()),d=f(get());
        if (a>b) swap(a,b);
        if (c>d) swap(c,d);
        printf("%d\n",ans=tr[d].qry(tr[c-1],a,b));
    }
    return 0;
}