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;
int n, m, s, num[100100], ck, ans;
pair<int, int> a, b;
struct Kuai {
  int cnt[100100], num[400 << 1], size;
  inline void add(int u, int v) {
    int i;
    cnt[v]++, size++;
    for (i = size; i > u; i--) num[i] = num[i - 1];
    num[u] = v;
  }
  inline void del(int u) {
    int i;
    cnt[num[u]]--, size--;
    for (i = u; i <= size; i++) num[i] = num[i + 1];
  }
  inline int ask(int u, int v, int w) {
    int i, res = 0;
    for (i = u; i <= v; i++) res += (num[i] == w);
    return res;
  }
} ku[400];
inline void build() {
  int i;
  for (i = 1; i <= n; i++) {
    ku[(i - 1) / s + 1].add((i - 1) % s + 1, num[i]);
  }
}
inline void rb() {
  int i, j, k = 0;
  for (i = 1; i <= ck; i++) {
    for (j = 1; j <= ku[i].size; j++) {
      num[++k] = ku[i].num[j];
      ku[i].cnt[num[k]] = 0;
    }
    ku[i].size = 0;
  }
  build();
}
int main() {
  int i, j, o, p, q, t, tmp;
  cin >> n;
  s = sqrt(n);
  ck = (n - 1) / s + 1;
  for (i = 1; i <= n; i++) scanf("%d", &num[i]);
  build();
  cin >> m;
  for (i = 1; i <= m; i++) {
    if (i % s == 0) rb();
    scanf("%d%d%d", &o, &p, &q);
    p = (p + ans - 1) % n + 1;
    q = (q + ans - 1) % n + 1;
    if (p > q) swap(p, q);
    if (o == 1) {
      if (p == q) continue;
      t = 0;
      for (j = 1;; j++) {
        if (t + ku[j].size >= q) break;
        t += ku[j].size;
      }
      tmp = ku[j].num[q - t];
      ku[j].del(q - t);
      t = 0;
      for (j = 1;; j++) {
        if (t + ku[j].size >= p) break;
        t += ku[j].size;
      }
      ku[j].add(p - t, tmp);
    } else {
      scanf("%d", &o);
      o = (o + ans - 1) % n + 1;
      ans = 0;
      t = 0;
      for (j = 1;; j++) {
        if (t + ku[j].size >= p) break;
        t += ku[j].size;
      }
      a = make_pair(j, p - t);
      for (;; j++) {
        if (t + ku[j].size >= q) break;
        t += ku[j].size;
      }
      b = make_pair(j, q - t);
      if (a.first == b.first)
        ans = ku[a.first].ask(a.second, b.second, o);
      else {
        for (j = a.first + 1; j < b.first; j++) ans += ku[j].cnt[o];
        ans += ku[a.first].ask(a.second, ku[a.first].size, o);
        ans += ku[b.first].ask(1, b.second, o);
      }
      printf("%d\n", ans);
    }
  }
}