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 a[100008], marks[500][100005], sexy_r, sexy_l, block;
queue<int> qu[500], q1;
void right(int x) {
  int y = (x - 1) % block + 1, u = (x - 1) / block + 1;
  int p = qu[u].size() - y;
  while (p--) {
    q1.push(qu[u].front());
    qu[u].pop();
  }
  sexy_r = qu[u].front();
  marks[u][sexy_r]--;
  qu[u].pop();
  while (!qu[u].empty()) {
    q1.push(qu[u].front());
    qu[u].pop();
  }
  while (!q1.empty()) {
    qu[u].push(q1.front());
    q1.pop();
  }
}
void left(int y, int op) {
  int o = (y - 1) % block + 1, u = (y - 1) / block + 1;
  int o2 = (op - 1) / block + 1;
  int p = qu[u].size() - o;
  while (p--) {
    q1.push(qu[u].front());
    qu[u].pop();
  }
  q1.push(qu[u].front());
  qu[u].pop();
  q1.push(sexy_r);
  marks[u][sexy_r]++;
  while (!qu[u].empty()) {
    q1.push(qu[u].front());
    qu[u].pop();
  }
  while (!q1.empty()) {
    qu[u].push(q1.front());
    q1.pop();
  }
}
int sol_rdel(int x, int z) {
  queue<int> q1;
  int y = (x - 1) % block + 1, u = (x - 1) / block + 1;
  int p = qu[u].size() - y;
  int coun = 0;
  while (p--) {
    q1.push(qu[u].front());
    if (qu[u].front() == z) coun++;
    qu[u].pop();
  }
  while (!qu[u].empty()) {
    q1.push(qu[u].front());
    qu[u].pop();
  }
  while (!q1.empty()) {
    qu[u].push(q1.front());
    q1.pop();
  }
  return coun;
}
int sol_ldel(int x, int z) {
  queue<int> q1;
  int y = (x - 1) % block + 1, u = (x - 1) / block + 1;
  int p = qu[u].size() - y;
  int coun = 0;
  while (p--) {
    q1.push(qu[u].front());
    qu[u].pop();
  }
  q1.push(qu[u].front());
  qu[u].pop();
  while (!qu[u].empty()) {
    q1.push(qu[u].front());
    if (qu[u].front() == z) coun++;
    qu[u].pop();
  }
  while (!q1.empty()) {
    qu[u].push(q1.front());
    q1.pop();
  }
  return coun;
}
int main() {
  int n, m, x, y, z;
  cin >> n;
  block = sqrt(n);
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = n; i >= 1; i--) {
    qu[(i - 1) / block + 1].push(a[i]);
    marks[(i - 1) / block + 1][a[i]]++;
  }
  cin >> m;
  int type, u, v, lans = 0, x1, y1, z1, o1, o2;
  while (m--) {
    cin >> type;
    if (type == 1) {
      cin >> x1 >> y1;
      o1 = ((x1 + lans - 1) % n) + 1;
      o2 = ((y1 + lans - 1) % n) + 1;
      x = min(o1, o2);
      y = max(o1, o2);
      u = (x - 1) / block + 1;
      v = (y - 1) / block + 1;
      if (x != y) {
        right(y);
        left(x, y);
      }
      for (int i = u; i < v; i++) {
        marks[i][qu[i].front()]--;
        marks[i + 1][qu[i].front()]++;
        qu[i + 1].push(qu[i].front());
        qu[i].pop();
      }
    } else {
      cin >> x1 >> y1 >> z1;
      o1 = ((x1 + lans - 1) % n) + 1;
      o2 = ((y1 + lans - 1) % n) + 1;
      z = ((z1 + lans - 1) % n) + 1;
      x = min(o1, o2);
      y = max(o1, o2);
      u = (x - 1) / block + 1;
      v = (y - 1) / block + 1;
      int sol = 0;
      for (int i = u; i <= v; i++) {
        sol += marks[i][z];
      }
      cout << sol - sol_rdel(y, z) - sol_ldel(x, z) << " " << endl;
      lans = sol - sol_rdel(y, z) - sol_ldel(x, z);
    }
  }
}