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 a[1000005], b[1000005], p[1000005], n;
struct seg {
  int l, r;
} t[4000005];
int sorted[1000005];
int tol[40][1000005];
int val[40][1000005];
void build(int l, int r, int i, int d) {
  t[i].l = l;
  t[i].r = r;
  if (l == r) return;
  int m = (l + r) >> 1;
  int lsame = m - l + 1;
  for (int k = l; k <= r; ++k)
    if (val[d][k] < sorted[m]) --lsame;
  int lpos = l, rpos = m + 1, same = 0;
  for (int k = l; k <= r; ++k) {
    if (k == l)
      tol[d][k] = 0;
    else
      tol[d][k] = tol[d][k - 1];
    if (val[d][k] < sorted[m])
      ++tol[d][k], val[d + 1][lpos++] = val[d][k];
    else if (val[d][k] > sorted[m])
      val[d + 1][rpos++] = val[d][k];
    else if (same < lsame)
      ++same, ++tol[d][k], val[d + 1][lpos++] = val[d][k];
    else
      val[d + 1][rpos++] = val[d][k];
  }
  build(l, m, i << 1, d + 1), build(m + 1, r, (i << 1) + 1, d + 1);
}
int find(int l, int r, int i, int d, int k) {
  if (l == r) return val[d][l];
  int s, ss;
  if (l == t[i].l)
    s = tol[d][r], ss = 0;
  else
    s = tol[d][r] - tol[d][l - 1], ss = tol[d][l - 1];
  if (s >= k) {
    int newl = t[i].l + ss;
    int newr = t[i].l + ss + s - 1;
    return find(newl, newr, i << 1, d + 1, k);
  } else {
    int m = (t[i].l + t[i].r) >> 1;
    int bb = l - t[i].l - ss;
    int b = r - l + 1 - s;
    int newl = m + bb + 1;
    int newr = m + bb + b;
    return find(newl, newr, (i << 1) + 1, d + 1, k - s);
  }
}
int cal(int fr, int k) {
  if (fr == 0) return 0;
  int fl = 1, l = 0, r = fr;
  while (l < r) {
    int m = (l + r + 1) >> 1;
    if (find(fl, fr, 1, 0, m) <= k)
      l = m;
    else
      r = m - 1;
  }
  return l;
}
int main() {
  int m, x = 0, la, ra, lb, rb;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
    p[a[i]] = i;
  }
  for (int i = 1; i <= n; ++i) {
    scanf("%d", b + i);
    sorted[i] = val[0][i] = p[b[i]];
  }
  sort(sorted + 1, sorted + n + 1);
  build(1, n, 1, 0);
  scanf("%d", &m);
  int cc = 0;
  while (m--) {
    scanf("%d%d%d%d", &la, &ra, &lb, &rb);
    la = la + x;
    while (la > n) la -= n;
    ra = ra + x;
    while (ra > n) ra -= n;
    lb = lb + x;
    while (lb > n) lb -= n;
    rb = rb + x;
    while (rb > n) rb -= n;
    if (la > ra) swap(la, ra);
    if (lb > rb) swap(lb, rb);
    int ll = cal(rb, la - 1) - cal(lb - 1, la - 1);
    int rr = cal(rb, ra) - cal(lb - 1, ra);
    x = rr - ll;
    ++x;
    ++cc;
    if (cc % 10000 == 0) cout << cc << endl;
  }
  return 0;
}
