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 = 1e6 + 5;
const int M = 2e5 + 5;
int rt[N];
int lson[N * 23], rson[N * 23], val[N * 23];
int n, m, ans, q;
int a[N], b[N], d[N];
void build(int pos, int x, int y) {
  if (x == y) return;
  int mid = (x + y) >> 1;
  lson[pos] = ++m;
  rson[pos] = ++m;
  build(lson[pos], x, mid);
  build(rson[pos], mid + 1, y);
}
void build(int ppos, int pos, int x, int y, int p) {
  if (x == y) {
    val[pos]++;
    return;
  }
  int mid = (x + y) >> 1;
  if (p <= mid) {
    lson[pos] = ++m;
    rson[pos] = rson[ppos];
    build(lson[ppos], lson[pos], x, mid, p);
    val[pos] = val[lson[pos]] + val[rson[pos]];
  } else {
    rson[pos] = ++m;
    lson[pos] = lson[ppos];
    build(rson[ppos], rson[pos], mid + 1, y, p);
    val[pos] = val[lson[pos]] + val[rson[pos]];
  }
}
int query(int pos, int x, int y, int l, int r) {
  if (x >= l && y <= r) return val[pos];
  if (x > r || y < l) return 0;
  int mid = (x + y) >> 1;
  return query(lson[pos], x, mid, l, r) + query(rson[pos], mid + 1, y, l, r);
}
int F(int x) { return (x - 1 + ans) % n + 1; }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  for (int i = 1; i <= n; i++) scanf("%d", b + i);
  for (int i = 1; i <= n; i++) d[b[i]] = i;
  rt[0] = ++m;
  build(rt[0], 1, n);
  for (int i = 1; i <= n; i++) {
    rt[i] = ++m;
    build(rt[i - 1], rt[i], 1, n, d[a[i]]);
  }
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int A, B, C, D, l1, r1, l2, r2;
    scanf("%d%d%d%d", &A, &B, &C, &D);
    l1 = min(F(A), F(B));
    r1 = max(F(A), F(B));
    l2 = min(F(C), F(D));
    r2 = max(F(C), F(D));
    ans = query(rt[r1], 1, n, l2, r2) - query(rt[l1 - 1], 1, n, l2, r2);
    printf("%d\n", ans);
    ans++;
  }
  return 0;
}