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 N = 1000010, D = N * 22;
int n, px[N], py[N], a[N], root[N];
int sum[D], l[D], r[D], dn;
int m, key, ans;
int insert(int d, int lt, int rt, int w) {
  int md = (lt + rt) >> 1;
  int nd = ++dn;
  sum[nd] = sum[d] + 1;
  if (lt != rt) {
    if (w <= md) {
      l[nd] = insert(l[d], lt, md, w);
      r[nd] = r[d];
    } else {
      r[nd] = insert(r[d], md + 1, rt, w);
      l[nd] = l[d];
    }
  }
  return nd;
}
int query(int d, int lt, int rt, int lq, int rq) {
  if (!d) {
    return 0;
  }
  if (lq <= lt && rt <= rq) {
    return sum[d];
  }
  int md = (lt + rt) >> 1;
  int ret = 0;
  if (lq <= md) {
    ret += query(l[d], lt, md, lq, rq);
  }
  if (md < rq) {
    ret += query(r[d], md + 1, rt, lq, rq);
  }
  return ret;
}
int init() {
  int x;
  scanf("%d", &x);
  return (x - 1 + key) % n + 1;
}
int main() {
  scanf("%d", &n);
  for (int i = 1, v; i <= n; ++i) {
    scanf("%d", &v), px[v] = i;
  }
  for (int i = 1, v; i <= n; ++i) {
    scanf("%d", &v), py[v] = i;
  }
  for (int v = 1; v <= n; ++v) {
    a[px[v]] = v;
  }
  root[0] = ++dn;
  for (int i = 1; i <= n; ++i) {
    root[i] = insert(root[i - 1], 1, n, py[a[i]]);
  }
  scanf("%d", &m);
  ans = -1;
  for (int im = 1; im <= m; ++im) {
    key = ans + 1;
    int l1 = init(), r1 = init(), l2 = init(), r2 = init();
    if (l1 > r1) swap(l1, r1);
    if (l2 > r2) swap(l2, r2);
    ans = query(root[r1], 1, n, l2, r2) - query(root[l1 - 1], 1, n, l2, r2);
    printf("%d\n", ans);
  }
}