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 N = 100005;
int main() {
  int n, q;
  int a[N];
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", a + i);
  int s = sqrt(n);
  deque<int> v[s + 2];
  int qtd[s + 2][n + 1];
  memset(qtd, 0, sizeof qtd);
  for (int i = 0; i < n; i++) {
    v[i / s].push_back(a[i]);
    qtd[i / s][a[i]]++;
  }
  int last = 0;
  scanf("%d", &q);
  while (q--) {
    int op, l, r, k;
    scanf("%d %d %d", &op, &l, &r);
    l = (l + last - 1) % n;
    r = (r + last - 1) % n;
    if (l > r) swap(l, r);
    vector<int> auxl, auxr;
    int ls = l / s;
    int rs = r / s;
    if (op == 1) {
      int ant = v[ls].back();
      for (int i = ls + 1; i < rs; i++) {
        qtd[i][ant]++;
        v[i].push_front(ant);
        ant = v[i].back();
        qtd[i][ant]--;
        v[i].pop_back();
      }
      if (rs != ls) {
        while (l / s == ls) {
          auxl.push_back(v[ls].back());
          v[ls].pop_back();
          l++;
        }
        reverse(auxl.begin(), auxl.end());
        while (r / s == rs) {
          auxr.push_back(v[rs].front());
          v[rs].pop_front();
          r--;
        }
        qtd[ls][*auxr.rbegin()]++;
        v[ls].push_back(*auxr.rbegin());
        qtd[ls][*auxl.rbegin()]--;
        for (int i = 0; i < auxl.size() - 1; i++) v[ls].push_back(auxl[i]);
        qtd[rs][*auxr.rbegin()]--;
        for (int i = auxr.size() - 2; i >= 0; i--) v[rs].push_front(auxr[i]);
        v[rs].push_front(ant);
        qtd[rs][ant]++;
      } else {
        while (v[ls].size()) {
          auxl.push_back(v[ls].front());
          v[ls].pop_front();
        }
        int ini = ls * s;
        l -= ini;
        r -= ini;
        ant = auxl[l];
        auxl[l] = auxl[r];
        for (int i = l + 1; i <= r; i++) swap(auxl[i], ant);
        for (int i = 0; i < auxl.size(); i++) v[ls].push_back(auxl[i]);
      }
    } else {
      int k;
      int ans = 0;
      scanf("%d", &k);
      k = ((k + last - 1) % n) + 1;
      for (int i = ls + 1; i < rs; i++) ans += qtd[i][k];
      if (ls != rs) {
        while (l / s == ls && l <= r) {
          if (v[ls].back() == k) ans++;
          auxl.push_back(v[ls].back());
          v[ls].pop_back();
          l++;
        }
        for (int i = auxl.size() - 1; i >= 0; i--) v[ls].push_back(auxl[i]);
        while (r / s == rs && r >= l) {
          if (v[rs].front() == k) ans++;
          auxr.push_back(v[rs].front());
          v[rs].pop_front();
          r--;
        }
        for (int i = auxr.size() - 1; i >= 0; i--) v[rs].push_front(auxr[i]);
      } else {
        while (v[ls].size()) {
          auxl.push_back(v[ls].front());
          v[ls].pop_front();
        }
        int ini = ls * s;
        l -= ini;
        r -= ini;
        for (int i = l; i <= r; i++) ans += auxl[i] == k;
        for (int i = 0; i < auxl.size(); i++) v[ls].push_back(auxl[i]);
      }
      last = ans;
      printf("%d\n", ans);
    }
  }
  return 0;
}