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<bits/stdc++.h>
using namespace std;
const int node_size = 1.1e6;
#define ls(o) tr[o].child[0]
#define rs(o) tr[o].child[1]
struct { int child[2],sum; } tr[node_size<<5];
int n,m,cnt,root[node_size];
int a[node_size],b[node_size],rev[node_size];
void modify(int lst,int &now,int l,int r,int pos)
{
    now=++cnt; tr[now]=tr[lst]; ++tr[now].sum;
    if(l==r) return void(); int mid=(l+r)>>1;
    if(pos<=mid) modify(ls(lst),ls(now),l,mid,pos);
        else     modify(rs(lst),rs(now),mid+1,r,pos);
}
int query(int lst,int now,int l,int r,int li,int ri)
{
    if(li<=l&&r<=ri) return tr[now].sum-tr[lst].sum;
    int mid=(l+r)>>1;
    if(ri<=mid) return query(ls(lst),ls(now),l,mid,li,ri);
    if(li >mid) return query(rs(lst),rs(now),mid+1,r,li,ri);
    return  query(ls(lst),ls(now),l,mid,li,ri)+
            query(rs(lst),rs(now),mid+1,r,li,ri);
}
void dfs(int now,int l,int r)
{
    if(l==r) return printf("%d%c",tr[now].sum," \n"[l==n]),void();
    int mid(l+r>>1); dfs(ls(now),l,mid),dfs(rs(now),mid+1,r);
}
int f(int lst,int val) { return (val+lst-1)%n+1; }
signed main()
{
#ifndef ONLINE_JUDGE
    freopen("test.in","r",stdin);
    freopen("test.out","w",stdout);
#endif
    scanf("%d",&n);
    for(int i=1;i<=n;++i) scanf("%d",a+i);
    for(int i=1;i<=n;++i) scanf("%d",b+i);
    for(int i=1;i<=n;++i) rev[a[i]]=i;
    for(int i=1;i<=n;++i) modify(root[i-1],root[i],1,n,rev[b[i]]);
    scanf("%d",&m);
    for(int a,b,c,d,lst=0;m;m--)
    {
        scanf("%d %d %d %d",&a,&b,&c,&d); lst%=n;
        a=f(lst,a),b=f(lst,b),c=f(lst,c),d=f(lst,d);
        if(a>b) swap(a,b); if(c>d) swap(c,d);
        int ans=query(root[c-1],root[d],1,n,a,b); lst+=ans;
        printf("%d\n",ans);
    }
    return 0;
}