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 long long INF = (long long)1e18;
const int mod = (int)1e9 + 7;
const int MAXN = (int)1e5 + 5;
long long n;
const int bs = 350;
int a[MAXN];
int mp[600][MAXN];
vector<int> ls[1000];
void swp(int l, int r) {
  int size = 0, i = 0;
  vector<int> tmp1, tmp2;
  for (i = 0; i <= (n - 1) / bs; i++) {
    if (size + ls[i].size() > r) {
      break;
    }
    size += ls[i].size();
  }
  long long sayi = 0;
  for (int j = 0; j < ls[i].size(); j++) {
    mp[i][ls[i][j]] = 0;
    if (j + size == r) {
      sayi = ls[i][j];
      continue;
    }
    tmp1.push_back(ls[i][j]);
  }
  ls[i].clear();
  for (int j : tmp1) {
    ls[i].push_back(j);
    mp[i][j]++;
  }
  size = 0;
  i = 0;
  for (i = 0; i <= (n - 1) / bs; i++) {
    if (size + ls[i].size() > l) {
      break;
    }
    size += ls[i].size();
  }
  if (l == n - 1) {
    ls[(l / bs)].push_back(sayi);
    mp[l / bs][sayi]++;
    return;
  }
  for (int j = 0; j < ls[i].size(); j++) {
    mp[i][ls[i][j]] = 0;
    if (j + size == l) {
      tmp2.push_back(sayi);
    }
    tmp2.push_back(ls[i][j]);
  }
  ls[i].clear();
  for (int j : tmp2) {
    ls[i].push_back(j);
    mp[i][j]++;
  }
}
long long get(long long ind, long long k) {
  if (ind < 0) return 0;
  long long ans = 0;
  int size = 0;
  int yer;
  for (int i = 0; i <= (n - 1) / bs; i++) {
    if (size + ls[i].size() > ind) {
      yer = i;
      break;
    }
    size += ls[i].size();
    ans += mp[i][k];
  }
  for (int i = 0; i + size <= ind; i++) {
    ans += (ls[yer][i] == k);
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    ls[i / bs].push_back(a[i]);
    mp[i / bs][a[i]]++;
  }
  int lastans = 0;
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    if (i % bs == 0) {
      vector<int> tmp;
      for (int i = 0; i <= (n - 1) / bs; i++) {
        for (auto j = ls[i].begin(); j < ls[i].end(); j++) {
          tmp.push_back(*j);
          mp[i][*j] = 0;
        }
      }
      for (int j = 0; j < n; j++) {
        if (j % bs == 0) {
          ls[j / bs].clear();
        }
        mp[j / bs][tmp[j]]++;
        ls[j / bs].push_back(tmp[j]);
      }
    }
    int a;
    cin >> a;
    if (a == 1) {
      int l, r;
      cin >> l >> r;
      l = (l + lastans - 1 + n) % n;
      r = (r + lastans - 1 + n) % n;
      if (l > r) swap(l, r);
      swp(l, r);
    } else {
      int l, r, k;
      cin >> l >> r >> k;
      l = (l + lastans - 1 + n) % n;
      r = (r + lastans - 1 + n) % n;
      k = 1 + (k + lastans - 1 + n) % n;
      if (l > r) swap(l, r);
      lastans = get(r, k) - get(l - 1, k);
      cout << lastans << '\n';
    }
  }
}