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 MAXN = 1e5 + 10;
const int BUCKET = 320;
int ans[MAXN][BUCKET], n;
deque<int> v[BUCKET];
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    v[i / BUCKET].push_front(x);
    ans[x][i / BUCKET]++;
  }
  int q;
  cin >> q;
  int ant = 0;
  for (int i = 1; i <= q; i++) {
    int op, l, r;
    cin >> op >> l >> r;
    if (op == 1) {
      l = (l + ant - 1) % n;
      r = (r + ant - 1) % n;
      if (l > r) swap(l, r);
      int k = l / BUCKET, p = r / BUCKET, x;
      vector<int> aux;
      int sz = v[p].size();
      for (int j = 0; j < sz; j++) {
        aux.push_back(v[p].back());
        ans[aux[j]][p]--;
        v[p].pop_back();
      }
      if (k < p) {
        v[p].push_front(v[p - 1].front());
        ans[v[p - 1].front()][p]++;
      }
      for (int j = 0; j < aux.size(); j++) {
        if (j == r - p * BUCKET) {
          x = aux[j];
          continue;
        }
        v[p].push_front(aux[j]);
        ans[aux[j]][p]++;
      }
      for (int j = p - 1; j > k; j--) {
        ans[v[j].front()][j]--;
        v[j].pop_front();
        ans[v[j - 1].front()][j]++;
        v[j].push_back(v[j - 1].front());
      }
      aux.clear();
      sz = v[k].size();
      for (int j = 0; j < sz; j++) {
        aux.push_back(v[k].back());
        ans[aux[j]][k]--;
        v[k].pop_back();
      }
      for (int j = 0; j < aux.size(); j++) {
        if (j == l - k * BUCKET) {
          v[k].push_front(x);
          ans[x][k]++;
        }
        if (k != p && j == aux.size() - 1) break;
        v[k].push_front(aux[j]);
        ans[aux[j]][k]++;
      }
    } else {
      int x;
      cin >> x;
      l = (l + ant - 1) % n;
      r = (r + ant - 1) % n;
      x = (x + ant - 1) % n + 1;
      if (l > r) swap(l, r);
      int k = l / BUCKET, p = r / BUCKET;
      if (k == p) {
        int resp = 0;
        vector<int> aux;
        int sz = v[k].size();
        for (int j = 0; j < sz; j++) {
          aux.push_back(v[k].back());
          v[k].pop_back();
          if (j < l - k * BUCKET || j > r - k * BUCKET) continue;
          if (aux[j] == x) resp++;
        }
        for (int j = 0; j < aux.size(); j++) v[k].push_front(aux[j]);
        cout << resp << "\n";
        ant = resp;
        continue;
      }
      int resp = 0;
      vector<int> aux;
      int sz = v[k].size();
      for (int j = 0; j < sz; j++) {
        aux.push_back(v[k].back());
        v[k].pop_back();
        if (j >= l - k * BUCKET && aux[j] == x) resp++;
      }
      for (int j = 0; j < aux.size(); j++) v[k].push_front(aux[j]);
      aux.clear();
      sz = v[p].size();
      for (int j = 0; j < sz; j++) {
        aux.push_back(v[p].back());
        v[p].pop_back();
        if (j <= r - p * BUCKET && aux[j] == x) resp++;
      }
      for (int j = 0; j < aux.size(); j++) v[p].push_front(aux[j]);
      for (int j = k + 1; j < p; j++) resp += ans[x][j];
      ant = resp;
      cout << resp << "\n";
    }
  }
}