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;
struct node {
  node *lc;
  node *rc;
  int cnt;
  node() : lc(0), rc(0), cnt(0){};
};
int n;
int a[1000005];
int pos[1000005];
node tree[10000000];
node *tot = tree;
node *root[1000005];
node *build(int l, int r) {
  node *ret = ++tot;
  if (r == l + 1) return ret;
  int mid = l + (r - l) / 2;
  ret->lc = build(l, mid);
  ret->rc = build(mid, r);
  return ret;
}
node *insert(node *const p, int l, int r, int val) {
  node *ret = ++tot;
  *ret = *p;
  ret->cnt += 1;
  if (val == l) return ret;
  int mid = l + (r - l) / 2;
  if (val < mid)
    ret->lc = insert(p->lc, l, mid, val);
  else
    ret->rc = insert(p->rc, mid, r, val);
  return ret;
}
int query(node *const beg, node *const end, int l, int r, int s, int t) {
  if (l == s && r == t) return end->cnt - beg->cnt;
  int mid = l + (r - l) / 2;
  int ans = 0;
  if (s < mid) ans += query(beg->lc, end->lc, l, mid, s, min(t, mid));
  if (t > mid) ans += query(beg->rc, end->rc, mid, r, max(s, mid), t);
  return ans;
}
int x = -1;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  for (int i = 1, j; i <= n; ++i) {
    scanf("%d", &j);
    pos[j] = i;
  }
  root[0] = build(1, n + 1);
  for (int i = 1; i <= n; ++i) {
    root[i] = insert(root[i - 1], 1, n + 1, pos[a[i]]);
  }
  int T;
  scanf("%d", &T);
  int v1, v2, v3, v4, l1, r1, l2, r2;
  while (T--) {
    scanf("%d%d%d%d", &v1, &v2, &v3, &v4);
    v1 = ((v1 + x + n) % n + 1), v2 = ((v2 + x + n) % n + 1),
    v3 = ((v3 + x + n) % n + 1), v4 = ((v4 + x + n) % n + 1);
    l1 = min(v1, v2), l2 = min(v3, v4);
    r1 = max(v1, v2), r2 = max(v3, v4);
    x = query(root[l1 - 1], root[r1], 1, n + 1, l2, r2 + 1);
    printf("%d\n", x);
  }
  return 0;
}
