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;
namespace IO {
const int sz = 1 << 15;
char inbuf[sz], outbuf[sz];
char *pinbuf = inbuf + sz;
char *poutbuf = outbuf;
inline char _getchar() {
  if (pinbuf == inbuf + sz) fread(inbuf, 1, sz, stdin), pinbuf = inbuf;
  return *(pinbuf++);
}
inline void _putchar(char x) {
  if (poutbuf == outbuf + sz) fwrite(outbuf, 1, sz, stdout), poutbuf = outbuf;
  *(poutbuf++) = x;
}
inline void flush() {
  if (poutbuf != outbuf)
    fwrite(outbuf, 1, poutbuf - outbuf, stdout), poutbuf = outbuf;
}
}  // namespace IO
inline int read() {
  int v = 0, f = 1;
  char c = IO::_getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = IO::_getchar();
  }
  while (c >= '0' && c <= '9') {
    v = v * 10 + c - '0';
    c = IO::_getchar();
  }
  return v * f;
}
const int Maxn = 1000005;
int n, p[Maxn], q[Maxn], P[Maxn], Q[Maxn], ord[Maxn];
vector<int> T[Maxn * 4];
void modify(int p, int l, int r, int x, int y) {
  T[p].push_back(y);
  if (l == r) return;
  int mid = l + r >> 1;
  if (x <= mid) {
    modify(p * 2, l, mid, x, y);
  } else {
    modify(p * 2 + 1, mid + 1, r, x, y);
  }
}
int query(int p, int l, int r, int a, int b, int c, int d) {
  if (a <= l && r <= b) {
    int pc = lower_bound(T[p].begin(), T[p].end(), c) - T[p].begin();
    int pd = upper_bound(T[p].begin(), T[p].end(), d) - T[p].begin();
    return pd - pc;
  }
  int mid = l + r >> 1;
  int ans = 0;
  if (a <= mid) {
    ans += query(p * 2, l, mid, a, b, c, d);
  }
  if (b > mid) {
    ans += query(p * 2 + 1, mid + 1, r, a, b, c, d);
  }
  return ans;
}
bool cmp(int x, int y) { return Q[x] < Q[y]; }
int main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    p[i] = read();
  }
  for (int i = 1; i <= n; i++) {
    P[p[i]] = i;
  }
  for (int i = 1; i <= n; i++) {
    q[i] = read();
  }
  for (int i = 1; i <= n; i++) {
    Q[q[i]] = i;
    ord[i] = i;
  }
  vector<pair<int, int> > V;
  sort(ord + 1, ord + 1 + n, cmp);
  for (int i = 1; i <= n; i++) {
    modify(1, 1, n, P[ord[i]], Q[ord[i]]);
  }
  int m = read();
  int lastans = 0;
  while (m--) {
    int a, b, c, d;
    a = read();
    b = read();
    c = read();
    d = read();
    a = (a - 1 + lastans) % n + 1;
    b = (b - 1 + lastans) % n + 1;
    c = (c - 1 + lastans) % n + 1;
    d = (d - 1 + lastans) % n + 1;
    int A = min(a, b), B = max(a, b), C = min(c, d), D = max(c, d);
    int rs = query(1, 1, n, A, B, C, D);
    lastans = rs + 1;
    printf("%d\n", rs);
  }
  return 0;
}