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 t = 1e5 + 5;
int a, b[t], c, s, e, l, r, k, s2, n, m, n2, m2, l2, r2, k2, cont;
struct ura {
  int cant[t], arr[700];
  int sz = 0;
} bkt[350];
void dest() {
  int pos = 0;
  for (int j = 0; j <= s2; j++) {
    for (int k = 0; k < bkt[j].sz; k++) {
      bkt[j].cant[bkt[j].arr[k]]--;
      b[pos++] = bkt[j].arr[k];
    }
  }
}
void cns() {
  int pos = 0;
  for (int i = 0; i < s2; i++) {
    for (int j = 0; j < s2; j++) {
      bkt[i].arr[j] = b[pos];
      bkt[i].cant[b[pos++]]++;
    }
    bkt[i].sz = s2;
  }
  int l = a - s2 * s2;
  for (int i = 0; i < l; i++) {
    bkt[s2].arr[i] = b[pos];
    bkt[s2].cant[b[pos++]]++;
  }
  bkt[s2].sz = l;
}
void quitar(int cub2) {
  bkt[cub2].cant[m]--;
  bkt[cub2].sz--;
  for (int i = n; i < bkt[cub2].sz; i++) {
    swap(bkt[cub2].arr[i], bkt[cub2].arr[i + 1]);
  }
}
void meter(int cub1) {
  bkt[cub1].cant[m]++;
  for (int i = bkt[cub1].sz - 1; n2 <= i; i--) {
    swap(bkt[cub1].arr[i], bkt[cub1].arr[i + 1]);
  }
  bkt[cub1].arr[n2] = m;
  bkt[cub1].sz++;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> a;
  s2 = sqrt(a);
  for (int i = 0; i < a; i++) cin >> b[i];
  cin >> c;
  s = sqrt(c);
  for (int i = 0; i < c; i++) {
    if (i % s == 0) {
      dest();
      cns();
    }
    cin >> e >> l2 >> r2;
    l = ((cont + l2 - 1) % a) + 1;
    r = ((cont + r2 - 1) % a) + 1;
    if (r < l) swap(l, r);
    int cub1 = 0, cub2 = 0, pos = 0;
    while (pos + bkt[cub1].sz < l) {
      pos += bkt[cub1].sz;
      cub1++;
    }
    n2 = l - pos - 1;
    m2 = bkt[cub1].arr[n2];
    cub2 = cub1;
    while (pos + bkt[cub2].sz < r) {
      pos += bkt[cub2].sz;
      cub2++;
    }
    n = r - pos - 1;
    m = bkt[cub2].arr[n];
    if (e == 1) {
      quitar(cub2);
      meter(cub1);
    } else {
      cin >> k2;
      k = ((cont + k2 - 1) % a) + 1;
      cont = 0;
      if (cub1 != cub2) {
        for (int j = n2; j < bkt[cub1].sz; j++)
          if (bkt[cub1].arr[j] == k) cont++;
        for (int j = cub1 + 1; j < cub2; j++) cont += bkt[j].cant[k];
        for (int j = 0; j <= n; j++)
          if (bkt[cub2].arr[j] == k) cont++;
      } else
        for (int j = n2; j <= n; j++)
          if (bkt[cub1].arr[j] == k) cont++;
      cout << cont << '\n';
    }
  }
  return 0;
}