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 sqt = 320;
int arr[300111];
vector<int> vec[320];
int rem[320][100001];
int c = 1;
pair<int, int> pos(int x) {
  pair<int, int> gerek;
  for (int i = 1; i <= c; i++) {
    if (vec[i].size() < x) {
      x -= vec[i].size();
    } else {
      gerek.first = i;
      for (int j = 0; j < vec[i].size(); j++) {
        if (j + 1 == x) {
          gerek.second = j;
          break;
        }
      }
      break;
    }
  }
  return gerek;
}
int query(int x, int y, int l, int r, int k) {
  int ans = 0;
  if (x == y) {
    for (int i = l; i <= r; i++) {
      if (vec[x][i] == k) ans++;
    }
    return ans;
  }
  for (int i = l; i < vec[x].size(); i++) {
    if (vec[x][i] == k) ans++;
  }
  for (int i = 0; i <= r; i++) {
    if (vec[y][i] == k) ans++;
  }
  for (int i = x + 1; i < y; i++) ans += rem[i][k];
  return ans;
}
void update(int x, int y, int l, int r) {
  int val;
  if (x == y) {
    val = vec[x][r];
    for (int i = r; i > l; i--) {
      vec[x][i] = vec[x][i - 1];
    }
    vec[x][l] = val;
    return;
  }
  val = vec[y][r];
  vec[x].push_back(0);
  for (int i = vec[x].size() - 1; i > l; i--) {
    vec[x][i] = vec[x][i - 1];
  }
  vec[x][l] = val;
  rem[x][val]++;
  rem[y][val]--;
  for (int i = r; i < vec[y].size() - 1; i++) {
    vec[y][i] = vec[y][i + 1];
  }
  vec[y].pop_back();
}
void reset() {
  vector<int> med;
  for (int i = 1; i <= c; i++) {
    for (int j = 0; j < vec[i].size(); j++) {
      med.push_back(vec[i][j]);
      rem[i][vec[i][j]] = 0;
    }
    vec[i].clear();
  }
  c = 1;
  for (int i = 0; i < med.size(); i++) {
    vec[c].push_back(med[i]);
    rem[c][med[i]]++;
    if ((i + 1) % sqt == 0) c++;
  }
}
int main() {
  int n, t;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &arr[i]);
  }
  for (int i = 1; i <= n; i++) {
    vec[c].push_back(arr[i]);
    rem[c][arr[i]]++;
    if (i % sqt == 0) c++;
  }
  scanf("%d", &t);
  int cnt = 0;
  int ans = 0;
  while (t--) {
    cnt++;
    if (cnt % sqt == 0) reset();
    int l, r, type;
    scanf("%d%d%d", &type, &l, &r);
    l = ((l + ans - 1) % n) + 1;
    r = ((r + ans - 1) % n) + 1;
    if (l > r) swap(l, r);
    pair<int, int> fup = pos(l);
    pair<int, int> tin = pos(r);
    if (type == 1) {
      update(fup.first, tin.first, fup.second, tin.second);
    } else {
      int k;
      scanf("%d", &k);
      k = ((k + ans - 1) % n) + 1;
      ans = query(fup.first, tin.first, fup.second, tin.second, k);
      printf("%d\n", ans);
    }
  }
  return 0;
}