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 = 2e6 + 10, M = 2e7 + 5;
int n, q, ans, cnt;
int ln[M], rn[M], sum[M];
int rt[N], a[N], pos[N];
template <class T>
inline void read(T &x) {
  x = 0;
  bool v = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') v = 1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + (c - '0');
    c = getchar();
  }
  if (v) x = -x;
}
inline int f(int x) { return (x + ans - 1) % n + 1; }
inline void defeat(int &cur, int l, int r, int x, int last) {
  cur = ++cnt;
  ln[cur] = ln[last];
  rn[cur] = rn[last];
  sum[cur] = sum[last] + 1;
  if (l == r) return;
  int mid = (l + r) >> 1;
  if (x <= mid)
    defeat(ln[cur], l, mid, x, ln[last]);
  else
    defeat(rn[cur], mid + 1, r, x, rn[last]);
}
inline int query(int cur, int l, int r, int L, int R) {
  if (L <= l && R >= r) {
    return sum[cur];
  }
  int res = 0;
  int mid = (l + r) >> 1;
  if (L <= mid) res += query(ln[cur], l, mid, L, R);
  if (R > mid) res += query(rn[cur], mid + 1, r, L, R);
  return res;
}
int main() {
  read(n);
  cout << n << endl;
  for (int i = 1; i <= n; i++) {
    int k;
    read(k);
    pos[k] = i;
  }
  for (int i = 1; i <= n; i++) {
    int k;
    read(k), a[i] = pos[k];
  }
  for (int i = 1; i <= n; i++) {
    defeat(rt[i], 1, n, a[i], rt[i - 1]);
  }
  read(q);
  while (q--) {
    int k;
    read(k);
    int x1 = f(k);
    read(k);
    int y1 = f(k);
    read(k);
    int x2 = f(k);
    read(k);
    int y2 = f(k);
    if (x1 > y1) swap(x1, y1);
    if (x2 > y2) swap(x2, y2);
    ans = query(rt[y2], 1, n, x1, y1) - query(rt[x2 - 1], 1, n, x1, y1);
    printf("%d\n", ans);
    ans++;
  }
  return 0;
}
