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;
template <class T>
inline void umax(T &a, T b) {
  if (a < b) a = b;
}
template <class T>
inline void umin(T &a, T b) {
  if (a > b) a = b;
}
template <class T>
inline T abs(T a) {
  return a > 0 ? a : -a;
}
template <class T>
inline T gcd(T a, T b) {
  return __gcd(a, b);
}
template <class T>
inline T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
const int inf = 1e9 + 143;
const long long longinf = 1e18 + 143;
inline int read() {
  int x;
  scanf(" %d", &x);
  return x;
}
const int N = 1010100;
int n;
int p[N], loca[N], locb[N];
vector<int> tree[N * 5];
void init(int x, int l, int r) {
  if (l == r) {
    tree[x].push_back(p[l]);
  } else {
    int m = (l + r) >> 1;
    init(x + x, l, m);
    init(x + x + 1, m + 1, r);
    tree[x].resize(tree[x + x].size() + tree[x + x + 1].size());
    merge((tree[x + x]).begin(), (tree[x + x]).end(), (tree[x + x + 1]).begin(),
          (tree[x + x + 1]).end(), tree[x].begin());
  }
}
int get(int x, int l, int r, int x1, int x2, int y1, int y2) {
  if (l > x2 || r < x1) return 0;
  if (l >= x1 && r <= x2)
    return upper_bound((tree[x]).begin(), (tree[x]).end(), y2) -
           lower_bound((tree[x]).begin(), (tree[x]).end(), y1);
  int m = (l + r) >> 1;
  return get(x + x, l, m, x1, x2, y1, y2) +
         get(x + x + 1, m + 1, r, x1, x2, y1, y2);
}
inline int f(int z, int x) { return (z + x - 1 + n) % n + 1; }
int main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    int x = read();
    loca[x] = i;
  }
  for (int i = 1; i <= n; i++) {
    int x = read();
    locb[x] = i;
  }
  for (int i = 1; i <= n; i++) {
    p[loca[i]] = locb[i];
  }
  init(1, 1, n);
  int q = read();
  int x = 0;
  while (q--) {
    int a = read(), b = read(), c = read(), d = read();
    int l1 = min(f(a, x), f(b, x));
    int r1 = max(f(a, x), f(b, x));
    int l2 = min(f(c, x), f(d, x));
    int r2 = max(f(c, x), f(d, x));
    0;
    x = get(1, 1, n, l1, r1, l2, r2);
    printf("%d\n", x);
    ++x;
  }
  return 0;
}