Problem: Serega loves fun. However, everyone has fun in the unique manner. Serega has fun by solving query problems. One day Fedor came up with such a problem.

You are given an array a consisting of n positive integers and queries to it. The queries can be of two types:

  1. Make a unit cyclic shift to the right on the segment from l to r (both borders inclusive). That is rearrange elements of the array in the following manner:a[l], a[l + 1], ..., a[r - 1], a[r] → a[r], a[l], a[l + 1], ..., a[r - 1].
  2. Count how many numbers equal to k are on the segment from l to r (both borders inclusive). 



Fedor hurried to see Serega enjoy the problem and Serega solved it really quickly. Let's see, can you solve it?

Input

The first line contains integer n (1 ≤ n ≤ 105) — the number of elements of the array. The second line contains n integers a[1], a[2], ..., a[n] (1 ≤ a[i] ≤ n).

The third line contains a single integer q (1 ≤ q ≤ 105) — the number of queries. The next q lines contain the queries.

As you need to respond to the queries online, the queries will be encoded. A query of the first type will be given in format: 1 l'i r'i. A query of the second type will be given in format: 2 l'i r'i k'i. All the number in input are integer. They satisfy the constraints: 1 ≤ l'i, r'i, k'i ≤ n.

To decode the queries from the data given in input, you need to perform the following transformations:

li = ((l'i + lastans - 1) mod n) + 1; ri = ((r'i + lastans - 1) mod n) + 1; ki = ((k'i + lastans - 1) mod n) + 1.

Where lastans is the last reply to the query of the 2-nd type (initially, lastans = 0). If after transformation li is greater than ri, you must swap these values.

Output

For each query of the 2-nd type print the answer on a single line.

Examples

Input

7
6 6 2 7 4 2 5
7
1 3 6
2 2 4 2
2 2 4 7
2 2 2 5
1 2 6
1 1 4
2 1 7 3


Output

2
1
0
0


Input

8
8 4 2 2 7 7 8 8
8
1 8 8
2 8 1 7
1 8 1
1 7 3
2 8 8 3
1 1 4
1 2 7
1 4 5


Output

2
0

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9 + 7;
struct cindylist {
  int data;
  cindylist *prev, *next;
  cindylist *n317;
  cindylist(int d = 0) {
    data = d;
    prev = next = n317 = NULL;
  }
  cindylist(const cindylist &oth) {
    data = oth.data;
    prev = oth.prev;
    next = oth.next;
    n317 = oth.n317;
  }
};
int cnt[317][100005];
cindylist *ptr[317];
vector<int> H;
cindylist *head;
void CL_init() { head = new cindylist; }
cindylist *CL_kth(int k) {
  if (k < 0) return head;
  cindylist *cur = head->next;
  while (k >= 317 && cur != NULL) {
    cur = cur->n317;
    k -= 317;
  }
  while (k > 0 && cur != NULL) {
    cur = cur->next;
    k--;
  }
  return cur;
}
int CL_count(int k, int v) {
  int ans = 0, cc = 0;
  cindylist *cur = head->next;
  while (k >= 317) {
    ans += cnt[cc][v];
    cc++;
    cur = cur->n317;
    k -= 317;
  }
  while (k > 0) {
    if (cur->data == v) ans++;
    cur = cur->next;
    k--;
  }
  return ans;
}
cindylist *CL_insert(cindylist *replc, cindylist *cont) {
  cont->next = replc->next;
  cont->prev = replc;
  if (replc->next != NULL) replc->next->prev = cont;
  replc->next = cont;
  cont->n317 = replc->n317;
  cindylist *cur = cont, *c2 = cont;
  for (int k = 0; k < (317); k++) {
    if (cur->prev != NULL)
      cur = cur->prev;
    else if (c2 != NULL)
      c2 = c2->next;
  }
  while (cur != cont) {
    cur->n317 = c2;
    cur = cur->next;
    if (c2 != NULL) c2 = c2->next;
  }
  cur = cont;
  for (int k = 0; k < (317); k++) {
    cur = cur->next;
    if (cur == NULL) break;
  }
  return cont;
}
void CL_erase(cindylist *replc) {
  replc->prev->next = replc->next;
  if (replc->next != NULL) replc->next->prev = replc->prev;
  cindylist *cur = replc->prev, *c2 = cur->next, *c3 = cur->next;
  for (int k = 0; k < (316); k++) {
    if (cur->prev != NULL)
      cur = cur->prev;
    else if (c2 != NULL)
      c2 = c2->next;
  }
  while (cur != c3) {
    cur->n317 = c2;
    cur = cur->next;
    if (c2 != NULL) c2 = c2->next;
  }
}
int main() {
  CL_init();
  cindylist *rear = head;
  int n, q, x, ans = 0;
  scanf("%d", &n);
  for (int k = 0; k < (n); k++) {
    scanf("%d", &x);
    rear = CL_insert(rear, new cindylist(x));
  }
  cindylist *cur = head->next;
  q = 0;
  while (cur != NULL) {
    cindylist *c2 = cur;
    for (int k = 0; k < (317); k++) {
      cnt[q][c2->data]++;
      c2 = c2->next;
      if (c2 == NULL) break;
    }
    ptr[q++] = cur;
    cur = cur->n317;
  }
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int t, v, l, r;
    scanf("%d%d%d", &t, &l, &r);
    l = (l + ans - 1) % n;
    r = (r + ans - 1) % n;
    if (l > r) swap(l, r);
    if (t == 1) {
      cindylist *mov = CL_kth(r), *cur;
      int E = r / 317;
      if (r % 317) {
        cur = ptr[E];
        cnt[E][mov->data]--;
        cur = cur->n317;
        if (cur != NULL) cnt[E][cur->data]++;
      } else {
        cur = mov;
        E--;
      }
      for (; cur != NULL; cur = cur->n317) {
        E++;
        ptr[E] = cur->next;
        cnt[E][cur->data]--;
        if (cur->n317 != NULL) cnt[E][cur->n317->data]++;
      }
      CL_erase(mov);
      mov = new cindylist(mov->data);
      CL_insert(CL_kth(l - 1), mov);
      E = l / 317;
      if (l % 317) {
        cur = ptr[E];
        cnt[E][mov->data]++;
        cur = cur->n317;
        if (cur != NULL) cnt[E][cur->data]--;
      } else {
        cur = mov;
        E--;
      }
      for (; cur != NULL; cur = cur->n317) {
        E++;
        ptr[E] = cur;
        cnt[E][cur->data]++;
        if (cur->n317 != NULL) cnt[E][cur->n317->data]--;
      }
    } else {
      scanf("%d", &v);
      v = (v + ans - 1) % n + 1;
      ans = CL_count(r + 1, v) - CL_count(l, v);
      printf("%d\n", ans);
    }
  }
  return 0;
}