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 N = 1000010;
int n, m, i, j, k, l, ans, l1, l2, r1, r2;
int P1[N], P2[N], Q1[N], Q2[N];
int f(int z) { return ((z - 1 + ans) % n) + 1; }
struct Functional_Segment_Tree {
  int tot, root[N];
  struct node {
    int l, r, size, son[2];
  } a[N * 21];
  void update(int x) {
    a[x].size = 0;
    if (a[x].son[0]) a[x].size += a[a[x].son[0]].size;
    if (a[x].son[1]) a[x].size += a[a[x].son[1]].size;
  }
  void build() {
    tot++;
    a[tot].l = 1;
    a[tot].r = n;
    root[0] = tot;
  }
  void insert(int x, int c) {
    if (a[x].l == a[x].r)
      a[x].size++;
    else {
      int mid = a[x].l + a[x].r >> 1;
      if (c <= mid) {
        a[++tot] = a[a[x].son[0]];
        a[x].son[0] = tot;
        a[tot].l = a[x].l;
        a[tot].r = mid;
        insert(tot, c);
      } else {
        a[++tot] = a[a[x].son[1]];
        a[x].son[1] = tot;
        a[tot].l = mid + 1;
        a[tot].r = a[x].r;
        insert(tot, c);
      }
      update(x);
    }
  }
  int search(int x, int l, int r) {
    if (a[x].l >= l && a[x].r <= r) return a[x].size;
    int mid = a[x].l + a[x].r >> 1;
    int re = 0;
    if (l <= mid && a[x].son[0]) re += search(a[x].son[0], l, r);
    if (r > mid && a[x].son[1]) re += search(a[x].son[1], l, r);
    return re;
  }
  void draw() {
    printf("root:");
    for (int i = 0; i <= n; i++) cout << root[i] << ' ';
    cout << endl;
    for (int i = 1; i <= tot; i++) {
      printf("node:%d len(%d,%d) data:%d son(%d,%d)\n", i, a[i].l, a[i].r,
             a[i].size, a[i].son[0], a[i].son[1]);
    }
  }
} tree;
int cmp1(int x, int y) { return P1[x] < P1[y]; }
int cmp2(int x, int y) { return P2[x] < P2[y]; }
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &j);
    Q1[j] = i;
  }
  for (i = 1; i <= n; i++) {
    scanf("%d", &j);
    Q2[j] = i;
  }
  for (i = 1; i <= n; i++) P1[Q1[i]] = Q2[i];
  tree.build();
  for (i = 1; i <= n; i++) {
    if (i % 100 == 0) cout << tree.tot << endl;
    tree.a[tree.root[i] = ++tree.tot] = tree.a[tree.root[i - 1]];
    tree.insert(tree.tot, P1[i]);
  }
  scanf("%d", &m);
  while (m--) {
    scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
    l1 = f(l1);
    r1 = f(r1);
    l2 = f(l2);
    r2 = f(r2);
    if (l1 > r1) swap(l1, r1);
    if (l2 > r2) swap(l2, r2);
    ans = tree.search(tree.root[r1], l2, r2) -
          tree.search(tree.root[l1 - 1], l2, r2);
    printf("%d\n", ans);
    ans++;
  }
  return 0;
}
