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, q;
int m;
int lastans;
int a[1000][700];
int b[100100];
int sz[1000];
short int cnt[1000][100100];
void remake() {
  int pos = 0;
  for (int i = 0; i <= m; i++) {
    sz[i] = 0;
    for (int j = 0; j < 700; a[i][j] = 0, j++) {
      if (a[i][j] == -1 || a[i][j] == 0) continue;
      cnt[i][a[i][j]]--;
      b[pos] = a[i][j];
      pos++;
    }
  }
  for (int i = 0; i < n; i++) {
    a[i / 333][i % 333] = b[i];
    cnt[i / 333][b[i]]++;
    sz[i / 333]++;
  }
}
int find(int it, int pos) {
  int i = 0;
  pos--;
  while (pos || a[it][i] == -1) {
    if (a[it][i] == -1) pos++;
    pos--;
    i++;
  }
  int res = a[it][i];
  a[it][i] = -1;
  return res;
}
void update(int it, int pos, int k) {
  pos--;
  int t = sz[it] - pos;
  int i = 0;
  while (pos || a[it][i] == -1) {
    if (a[it][i] == -1) pos++;
    pos--;
    i++;
  }
  int buf = a[it][i];
  a[it][i] = k;
  while (t) {
    i++;
    if (buf != -1) t--;
    swap(a[it][i], buf);
  }
}
int solve(int r, int k) {
  int res = 0;
  int pos = 0;
  int it = 0;
  while (it <= m && sz[it] <= r) {
    r -= sz[it];
    res += cnt[it][k];
    it++;
  }
  int i = 0;
  while (pos < r) {
    if (a[it][i] == -1 || a[it][i] == 0) {
      i++;
      continue;
    }
    res += (a[it][i] == k);
    i++;
    pos++;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i / 333][i % 333];
    cnt[i / 333][a[i / 333][i % 333]]++;
    sz[i / 333]++;
  }
  m = (n - 1) / 333 + 1;
  cin >> q;
  while (q--) {
    int type, l, r, k;
    cin >> type >> l >> r;
    l = (l + lastans - 1) % n + 1;
    r = (r + lastans - 1) % n + 1;
    if (l > r) swap(l, r);
    if (type == 1) {
      int it = 0;
      int pos = 0;
      while (pos < l) {
        pos += sz[it];
        it++;
      }
      if (pos >= l) {
        it--;
        pos -= sz[it];
      }
      int p1 = pos;
      int w = it;
      it = 0;
      pos = 0;
      while (pos < r) {
        pos += sz[it];
        it++;
      }
      if (pos >= r) {
        it--;
        pos -= sz[it];
      }
      k = find(it, r - pos);
      cnt[w][k]++;
      cnt[it][k]--;
      sz[w]++;
      sz[it]--;
      update(w, l - p1, k);
      if (sz[w] > 2 * 333) remake();
    } else {
      cin >> k;
      k = (k + lastans - 1) % n + 1;
      lastans = solve(r, k) - solve(l - 1, k);
      printf("%d\n", lastans);
    }
  }
}