Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MOD = 998244353;
struct tnode {
  int sum;
  tnode *lson, *rson;
  tnode(int x = 0) {
    sum = x;
    lson = rson = NULL;
  }
};
void pushup(tnode* cur) {
  cur->sum = (cur->lson == NULL ? 0 : cur->lson->sum) +
             (cur->rson == NULL ? 0 : cur->rson->sum);
}
tnode* modify(tnode* cur, int id, int val, int cl = 0, int cr = 1048575) {
  if (cl == cr) return new tnode(val);
  int mid = (cl + cr) >> 1;
  tnode* ret = new tnode();
  tnode *ls = cur == NULL ? NULL : cur->lson,
        *rs = cur == NULL ? NULL : cur->rson;
  ret->lson = id <= mid ? modify(ls, id, val, cl, mid) : ls;
  ret->rson = id > mid ? modify(rs, id, val, mid + 1, cr) : rs;
  pushup(ret);
  return ret;
}
int query(tnode* cur, int l, int r, int cl = 0, int cr = 1048575) {
  if (cur == NULL) return 0;
  if (l == cl && r == cr) return cur->sum;
  int mid = (cl + cr) >> 1;
  if (r <= mid)
    return query(cur->lson, l, r, cl, mid);
  else if (l > mid)
    return query(cur->rson, l, r, mid + 1, cr);
  else
    return query(cur->lson, l, mid, cl, mid) +
           query(cur->rson, mid + 1, r, mid + 1, cr);
}
int n, q, p0[1000005], occ[1000005], p1[1000005];
tnode* tre[1000005];
int x;
int f(int z) { return (z - 1 + x) % n + 1; }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= (int)(n); i++) {
    scanf("%d", &p0[i]);
    occ[p0[i]] = i;
  }
  for (int i = 1; i <= (int)(n); i++) scanf("%d", &p1[i]);
  for (int i = 1; i <= (int)(n); i++)
    tre[i] = modify(tre[i - 1], occ[p1[i]], 1);
  scanf("%d", &q);
  x = 0;
  for (int i = 0; i < (int)(q); i++) {
    int a, b, c, d;
    scanf("%d%d%d%d", &a, &b, &c, &d);
    a = f(a);
    b = f(b);
    c = f(c);
    d = f(d);
    if (a > b) swap(a, b);
    if (c > d) swap(c, d);
    x = query(tre[d], a, b) - query(tre[c - 1], a, b) + 1;
    printf("%d\n", x - 1);
  }
  return 0;
}