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
1
1
1
1
1 1 1 1

1

#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0;
  char ch = getchar();
  while (!isdigit(ch)) ch = getchar();
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return x;
}
int pos[200010];
unsigned long long Pow[200010];
struct SegmentTree {
  unsigned long long val[200010 << 2];
  int num[200010 << 2];
  inline void update(int k) {
    num[k] = num[k << 1] + num[k << 1 | 1];
    val[k] = val[k << 1] * Pow[num[k << 1 | 1]] + val[k << 1 | 1];
  }
  inline void change(int k, int l, int r, int p, int dat) {
    if (p < l || r < p) return;
    if (l == r) return (void)(val[k] = dat, num[k] = (dat != 0));
    change(k << 1, l, ((l + r) >> 1), p, dat),
        change(k << 1 | 1, ((l + r) >> 1) + 1, r, p, dat), update(k);
  }
} T;
int main() {
  int n = read(), m = read();
  unsigned long long H = 0, sum = 0;
  Pow[0] = 1;
  for (int i = 1; i <= n; i++) {
    int x = read();
    H = H * 23 + x;
    Pow[i] = Pow[i - 1] * 23;
    sum += Pow[i - 1];
  }
  int ans = 0;
  for (int i = 1; i <= m; i++) pos[read()] = i;
  for (int i = 1; i <= m; i++) {
    if (i > n) T.change(1, 1, m, pos[i - n], 0);
    T.change(1, 1, m, pos[i], i);
    if (i >= n && T.val[1] == 1ULL * (i - n) * sum + H) ++ans;
  }
  cout << ans << endl;
  return 0;
}
