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 = (int)1e5 + 7;
int n, q, a[N], BLOCK_SIZE, nblocks;
struct block {
  vector<int> cnt;
  deque<int> elem;
  block() {
    cnt.clear();
    cnt.resize(N);
    elem.clear();
  }
};
vector<block> lsts;
pair<int, int> get(int l) {
  int idx = l / BLOCK_SIZE;
  int startPos = l % BLOCK_SIZE;
  return make_pair(idx, startPos);
}
int half_query(int idx, int l, int r, int k) {
  int ans = 0;
  for (int i = l; i <= r; i++) {
    if (lsts[idx].elem[i] == k) {
      ans++;
    }
  }
  return ans;
}
int query(int l, int r, int k) {
  pair<int, int> p1 = get(l);
  pair<int, int> p2 = get(r);
  int ans = 0;
  if (p1.first == p2.first) {
    ans = half_query(p1.first, p1.second, p2.second, k);
  } else {
    ans +=
        half_query(p1.first, p1.second, (int)lsts[p1.first].elem.size() - 1, k);
    ans += half_query(p2.first, 0, p2.second, k);
    for (int i = p1.first + 1; i < p2.first; i++) {
      ans += lsts[i].cnt[k];
    }
  }
  return ans;
}
vector<int> unupdated_deque;
int margin;
void complete_update(int idx) {
  int prev = unupdated_deque[idx - 1 - margin];
  lsts[idx].cnt[prev]++;
  lsts[idx].elem.push_front(prev);
  int toRemove = unupdated_deque[idx - margin];
  lsts[idx].elem.pop_back();
  lsts[idx].cnt[toRemove]--;
}
void print(deque<int> d) {
  for (int i = 0; i < d.size(); i++) cout << d[i] << " ";
  cout << endl;
}
void printAll() {
  cout << "entire block is being printed" << endl;
  for (int i = 0; i < lsts.size(); i++) {
    print(lsts[i].elem);
  }
}
void update(int l, int r) {
  pair<int, int> p1 = get(l);
  pair<int, int> p2 = get(r);
  if (p1.first == p2.first) {
    {
      int idx = p1.first;
      int last = lsts[idx].elem[p2.second];
      for (int i = p1.second; i <= p2.second; i++) {
        int new_last = lsts[idx].elem[i];
        lsts[idx].elem[i] = last;
        last = new_last;
      }
    }
  } else {
    unupdated_deque.clear();
    for (int i = p1.first; i <= p2.first; i++) {
      unupdated_deque.push_back(lsts[i].elem.back());
    }
    margin = p1.first;
    {
      int idx = p1.first;
      int last = lsts[p2.first].elem[p2.second];
      int toRemove = lsts[idx].elem.back();
      lsts[idx].cnt[toRemove]--;
      int toAdd = last;
      lsts[idx].cnt[toAdd]++;
      for (int i = p1.second; i < lsts[idx].elem.size(); i++) {
        int new_last = lsts[idx].elem[i];
        lsts[idx].elem[i] = last;
        last = new_last;
      }
    }
    {
      int idx = p2.first;
      assert(idx > 0);
      int last = unupdated_deque[idx - 1 - margin];
      int toAdd = last;
      lsts[idx].cnt[toAdd]++;
      int toRemove = lsts[p2.first].elem[p2.second];
      lsts[idx].cnt[toRemove]--;
      for (int i = 0; i <= p2.second; i++) {
        int new_last = lsts[idx].elem[i];
        lsts[idx].elem[i] = last;
        last = new_last;
      }
    }
    for (int i = p1.first + 1; i < p2.first; i++) {
      complete_update(i);
    }
  }
}
void insert(vector<int> v) {
  block my_block;
  for (int i = 0; i < v.size(); i++) {
    my_block.cnt[v[i]]++;
    my_block.elem.push_back(v[i]);
  }
  lsts.push_back(my_block);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  BLOCK_SIZE = sqrt(n) + 1;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<int> temp;
  for (int i = 0; i < n; i++) {
    if (i > 0 && i % BLOCK_SIZE == 0) {
      insert(temp);
      temp.clear();
      temp.push_back(a[i]);
    } else {
      temp.push_back(a[i]);
    }
  }
  if (temp.size()) {
    insert(temp);
  }
  cin >> q;
  int lastans = 0;
  int ct = 0;
  while (q--) {
    int type;
    cin >> type;
    if (type == 1) {
      int l1, r1;
      cin >> l1 >> r1;
      int l = (l1 + lastans - 1) % n + 1;
      int r = (r1 + lastans - 1) % n + 1;
      if (l > r) swap(l, r);
      update(l - 1, r - 1);
    } else {
      int l1, r1, k1;
      cin >> l1 >> r1 >> k1;
      int l = (l1 + lastans - 1) % n + 1;
      int r = (r1 + lastans - 1) % n + 1;
      int k = (k1 + lastans - 1) % n + 1;
      l--, r--;
      if (l > r) swap(l, r);
      int ans = query(l, r, k);
      lastans = ans;
      cout << ans << endl;
    }
  }
  return 0;
}