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>
#pragma comment(linker, "/STACK:1000000000,1000000000")
using namespace std;
const long long inf = 1e18 + 7;
const long long mod = 1e9 + 7;
const double eps = 1e-9;
const double PI = 2 * acos(0.0);
const double E = 2.71828;
long long n, m, ans = 0;
long long a[1000005], b[1000005];
vector<long long> t[4000005];
inline long long f(long long x) {
  long long res = ((x + ans - 1) % n + n) % n;
  return res;
}
inline void build(long long v, long long l, long long r) {
  if (l == r) {
    t[v].push_back(a[l]);
    return;
  }
  long long m = (l + r) >> 1LL;
  if (l <= m) build(v + v, l, m);
  if (r > m) build(v + v + 1, m + 1, r);
  merge(t[v + v].begin(), t[v + v].end(), t[v + v + 1].begin(),
        t[v + v + 1].end(), back_inserter(t[v]));
}
inline long long get(long long v, long long tl, long long tr, long long a,
                     long long b, long long c, long long d) {
  if (a <= tl && b >= tr) {
    return upper_bound((t[v]).begin(), (t[v]).end(), d) -
           lower_bound((t[v]).begin(), (t[v]).end(), c);
  }
  long long tm = (tl + tr) >> 1LL;
  long long res = 0;
  if (a <= tm) res += get(v + v, tl, tm, a, b, c, d);
  if (b > tm) res += get(v + v + 1, tm + 1, tr, a, b, c, d);
  return res;
}
int main(void) {
  cin >> n;
  for (long long(i) = 0; (i) < (long long)(n); (i)++) {
    int x;
    scanf("%d", &x);
    b[x - 1] = i;
  }
  for (long long(i) = 0; (i) < (long long)(n); (i)++) {
    int x;
    scanf("%d", &x);
    a[i] = b[x - 1];
  }
  build(1, 0, n - 1);
  cin >> m;
  for (long long(i) = 0; (i) < (long long)(m); (i)++) {
    int a, b, c, d;
    scanf("%d%d%d%d", &a, &b, &c, &d);
    a = f(a);
    b = f(b);
    c = f(c);
    d = f(d);
    if (a > b) swap(a, b);
    if (c > d) swap(c, d);
    ans = get(1, 0, n - 1, c, d, a, b);
    printf("%d\n", (int)ans);
    ans++;
  }
  return 0;
}