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 = 1e6 + 5, M = 21 * N;
int n, pos[N], a[N], b[N], rt[N], m, l, r, x, y, ans;
struct zxs {
  int tot, ls[M], rs[M], sum[M];
  inline void pushup(int u) { sum[u] = sum[ls[u]] + sum[rs[u]]; }
  inline void update(int &u, int pre, int l, int r, int x) {
    u = ++tot;
    if (l == r) {
      sum[u]++;
      return;
    } else {
      int mid = (l + r) >> 1;
      ls[u] = ls[pre], rs[u] = rs[pre];
      if (x <= mid) {
        update(ls[u], ls[pre], l, mid, x);
      } else {
        update(rs[u], rs[pre], mid + 1, r, x);
      }
      pushup(u);
    }
  }
  inline int query(int u, int l, int r, int x, int y) {
    if (x <= l && y >= r)
      return sum[u];
    else {
      int val = 0;
      int mid = (l + r) >> 1;
      if (x <= mid) val += query(ls[u], l, mid, x, y);
      if (y > mid) val += query(rs[u], mid + 1, r, x, y);
      return val;
    }
  }
} tree;
inline int read() {
  int x;
  scanf("%d", &x);
  return (x - 1 + ans) % n + 1;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    int temp;
    scanf("%d", &temp);
    a[temp] = i;
  }
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &b[i]);
    pos[i] = a[b[i]];
  }
  for (int i = 1; i <= n; ++i) {
    tree.update(rt[i], rt[i - 1], 1, n, pos[i]);
  }
  scanf("%d", &m);
  for (int i = 1; i <= m; ++i) {
    l = read(), r = read(), x = read(), y = read();
    if (l > r) swap(l, r);
    if (x > y) swap(x, y);
    ans = tree.query(rt[y], 1, n, l, r);
    ans -= tree.query(rt[x - 1], 1, n, l, r);
    printf("%d\n", ans);
    ans++;
  }
  return 0;
}