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 = 1000005, M = 22000005;
int n, m, tot, ans;
int a[N], posb[N], rt[N], cnt[M], lc[M], rc[M];
void pushup(int u) { cnt[u] = cnt[lc[u]] + cnt[rc[u]]; }
void upd(int &u, int v, int l, int r, int x) {
  u = ++tot;
  if (l == r) {
    cnt[u] = cnt[v] + 1;
    return;
  }
  int mid = (l + r) >> 1;
  if (mid >= x)
    rc[u] = rc[v], upd(lc[u], lc[v], l, mid, x);
  else
    lc[u] = lc[v], upd(rc[u], rc[v], mid + 1, r, x);
  pushup(u);
}
int qsum(int u, int v, int l, int r, int ql, int qr) {
  if (l >= ql && r <= qr) return cnt[v] - cnt[u];
  int mid = (l + r) >> 1, res = 0;
  if (lc[v] && mid >= ql) res += qsum(lc[u], lc[v], l, mid, ql, qr);
  if (rc[v] && mid < qr) res += qsum(rc[u], rc[v], mid + 1, r, ql, qr);
  return res;
}
inline int f(int x) { return (x + ans) % n + 1; }
inline int input() {
  int x = 0, F = 1;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') F = -1;
    c = getchar();
  }
  while (c <= '9' && c >= '0') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * F;
}
int main() {
  int A, B, C, D;
  n = input();
  for (int i = 1; i <= n; i++) a[i] = input();
  for (int i = 1; i <= n; i++) A = input(), posb[A] = i;
  for (int i = 1; i <= n; i++) upd(rt[i], rt[i - 1], 1, n, posb[a[i]]);
  m = input();
  ans = -1;
  for (int i = 1; i <= m; i++) {
    A = input();
    B = input();
    C = input();
    D = input();
    A = f(A);
    B = f(B);
    C = f(C);
    D = f(D);
    if (A > B) swap(A, B);
    if (C > D) swap(C, D);
    printf("%d\n", ans = qsum(rt[A - 1], rt[B], 1, n, C, D));
  }
  return 0;
}