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;
int tp = 1, ans = 1, n, q;
struct Node {
  int lc, rc;
  int lq, rq;
  int sum;
};
Node t[4 * 1010240 + 20 * 5010];
void bld(int k, int l, int r) {
  t[k].lq = l;
  t[k].rq = r;
  if (l == r) return;
  int m = (l + r) >> 1;
  t[k].lc = (++tp);
  bld(tp, l, m);
  t[k].rc = (++tp);
  bld(tp, m + 1, r);
  return;
}
void mdf(int k, int g, int x) {
  t[k] = t[g];
  int l = t[k].lq, r = t[k].rq, m = (l + r) >> 1;
  t[k].sum++;
  if (l == r) return;
  if (x <= m) {
    t[k].lc = (++tp);
    mdf(tp, t[g].lc, x);
  }
  if (x > m) {
    t[k].rc = (++tp);
    mdf(tp, t[g].rc, x);
  }
  return;
}
int qry(int f, int g, int v, int w) {
  int l = t[f].lq, r = t[f].rq;
  if (l >= v && r <= w) return t[g].sum - t[f].sum;
  if (l > w || r < v) return 0;
  return qry(t[f].lc, t[g].lc, v, w) + qry(t[f].rc, t[g].rc, v, w);
}
int fc(int z) { return ((z - ans) % n + n) % n + 1; }
int mn(int z, int w) {
  if (z < w) return z;
  return w;
}
int mx(int z, int w) {
  if (z > w) return z;
  return w;
}
int a[101024], f[101024], xc, rt[101024];
int main() {
  scanf("%d", &n);
  bld(1, 1, n);
  rt[0] = 1;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    f[a[i]] = i;
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &xc);
    rt[i] = ++tp;
    mdf(tp, rt[i - 1], f[xc]);
  }
  scanf("%d", &q);
  while (q--) {
    int q, w, e, r;
    scanf("%d%d%d%d", &q, &w, &e, &r);
    q = mn(fc(q), fc(w));
    w = mx(fc(q), fc(w));
    e = mn(fc(e), fc(r));
    r = mx(fc(e), fc(r));
    ans = qry(rt[e - 1], rt[r], q, w);
    printf("%d\n", ans);
  }
  return 0;
}
