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
#include <bits/stdc++.h>
using namespace std;
const int TAM = 100100;
const int SZ = 317;
struct Bucket {
  deque<int> deq;
  int cnt[TAM];
  inline int size() { return deq.size(); }
  inline int count(int fr, int to, int k) {
    if (fr < 0) fr = 0;
    if (to >= size()) to = size() - 1;
    if (fr == 0 && to + 1 == size()) return cnt[k];
    int ans = 0;
    for (int i = fr; i <= to; ++i)
      if (deq[i] == k) ans++;
    return ans;
  }
  inline void dec(int val) { cnt[val]--; }
  void insert(int i, int val) {
    cnt[val]++;
    deq.insert(deq.begin() + i, val);
  }
  void erase(int i) {
    dec(deq[i]);
    deq.erase(deq.begin() + i);
  }
  void pop_back() {
    dec(deq.back());
    deq.pop_back();
  }
  void push_back(int val) {
    deq.push_back(val);
    cnt[val]++;
  }
  void pop_front() {
    dec(deq.front());
    deq.pop_front();
  }
  void push_front(int val) {
    deq.push_front(val);
    cnt[val]++;
  }
  inline int front() { return deq.front(); }
  inline int back() { return deq.back(); }
};
vector<Bucket> a;
int counter = 0;
int query(int fr, int to, int k) {
  int lo = 0, ans = 0;
  for (int it = 0; it < int(a.size()); ++it) {
    ans += a[it].count(fr - lo, to - lo, k);
    lo += a[it].size();
  }
  return ans;
}
void propagate() {
  for (int i = 0, j = 0, x; i + 1 < int(a.size()); ++i) {
    if (i == j) j++;
    while (a[i].size() < SZ) {
      while (a[j].size() == 0) j++;
      x = a[j].front();
      a[j].pop_front();
      a[i].push_back(x);
    }
    while (a[i].size() > SZ) {
      x = a[i].back();
      a[i].pop_back();
      a[j].push_front(x);
    }
  }
}
void insert(int i, int x) {
  int lo = 0;
  for (int it = 0; it < int(a.size()); ++it) {
    if (lo <= i && i < lo + a[it].size()) {
      a[it].insert(i - lo, x);
    }
    lo += a[it].size();
  }
}
int erase(int i) {
  int lo = 0, ans;
  for (int it = 0; it < int(a.size()); ++it) {
    if (lo <= i && i < lo + a[it].size()) {
      ans = a[it].deq[i - lo];
      a[it].erase(i - lo);
      return ans;
    }
    lo += a[it].size();
  }
  return -1;
}
void update(int fr, int to) {
  if (fr == to) return;
  counter++;
  if (counter == SZ) {
    counter = 0;
    propagate();
  }
  int x = erase(to);
  insert(fr, x);
}
void print() {
  for (int i = 0; i < int(a.size()); ++i) {
    cout << " {";
    for (int j = 0; j < a[i].size(); ++j) cout << ' ' << a[i].deq[j];
    cout << " }";
  }
  cout << endl;
}
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  int lastans = 0;
  int n;
  cin >> n;
  a.assign(n / SZ + (n % SZ > 0 ? 1 : 0), Bucket());
  for (int i = 0, x; i < n; ++i) {
    cin >> x;
    a[i / SZ].push_back(x);
  }
  int nQueries;
  cin >> nQueries;
  int cmd, fr, to, k;
  while (nQueries--) {
    cin >> cmd >> fr >> to;
    fr = (fr + lastans + n - 1) % n;
    to = (to + lastans + n - 1) % n;
    if (fr > to) swap(fr, to);
    if (cmd == 1)
      update(fr, to);
    else {
      cin >> k;
      k = (k + lastans + n - 1) % n + 1;
      lastans = query(fr, to, k);
      cout << lastans << '\n';
    }
  }
  return 0;
}
