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>
static inline unsigned long long rdtsc() {
  unsigned long long d;
  __asm__ __volatile__("rdtsc" : "=A"(d));
  return d;
}
using namespace std;
const int inf = (int)1e9;
const double eps = 1e-9;
const int maxV = (int)23e6 + 10;
const int maxn = (int)1e6 + 10;
int t[maxV];
int l[maxV], r[maxV];
int root[maxn];
int Last = 0;
inline int init(int L, int R) {
  int v = Last++;
  if (L + 1 < R) {
    int M = (L + R) >> 1;
    l[v] = init(L, M);
    r[v] = init(M, R);
  }
  return v;
}
inline int NEW(int x) {
  assert(Last < maxV);
  t[Last] = t[x];
  l[Last] = l[x];
  r[Last] = r[x];
  return Last++;
}
int modify(int x, int L, int R, int d) {
  if (d >= R || L > d) return x;
  x = NEW(x);
  if (L + 1 == R) {
    t[x] = 1;
    return x;
  }
  int M = (L + R) >> 1;
  l[x] = modify(l[x], L, M, d);
  r[x] = modify(r[x], M, R, d);
  t[x] = t[l[x]] + t[r[x]];
  return x;
}
int n;
int p[maxn], q[maxn];
int q1[maxn];
void precalc() {
  for (int i = 0; (i) < (n); ++i) p[i] -= 1;
  for (int i = 0; (i) < (n); ++i) q[i] -= 1;
  for (int i = 0; (i) < (n); ++i) q1[q[i]] = i;
  root[0] = init(0, n);
  for (int i = 0; (i) < (n); ++i) {
    root[i + 1] = modify(root[i], 0, n, q1[p[i]]);
  }
}
inline int gget(int x, int L, int R, int lq, int rq) {
  if (lq <= L && R <= rq) return t[x];
  if (lq >= R || L >= rq) return 0;
  int M = (L + R) >> 1;
  return gget(l[x], L, M, lq, rq) + gget(r[x], M, R, lq, rq);
}
inline int get(int r1, int l2, int r2) {
  return gget(root[r1 + 1], 0, n, l2, r2 + 1);
}
inline int GET(int l1, int r1, int l2, int r2) {
  l1 -= 1;
  return get(r1, l2, r2) - get(l1, l2, r2);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; (i) < (n); ++i) scanf("%d", &p[i]);
  for (int i = 0; (i) < (n); ++i) scanf("%d", &q[i]);
  precalc();
  int m;
  scanf("%d", &m);
  int x = 0;
  for (int i = 0; (i) < (m); ++i) {
    int a, b, c, d;
    scanf("%d%d%d%d", &a, &b, &c, &d);
    int fa = ((a - 1 + x) % n);
    int fb = ((b - 1 + x) % n);
    int fc = ((c - 1 + x) % n);
    int fd = ((d - 1 + x) % n);
    int l1 = min(fa, fb);
    int r1 = max(fa, fb);
    int l2 = min(fc, fd);
    int r2 = max(fc, fd);
    int res = GET(l1, r1, l2, r2);
    printf("%d\n", res);
    x = res + 1;
  }
  return 0;
}