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 maxN = 100 * 1000 + 5;
const int mod = 1000 * 1000 * 1000 + 7;
const int SQ = 376;
int fen[SQ][maxN], a[maxN];
inline void add(int ps, int n, int k) { fen[ps][n] += k; }
inline int get(int ps, int n) { return fen[ps][n]; }
deque<int> d[SQ];
void shift(int l, int r) {
  int stl = l / SQ;
  int str = r / SQ;
  if (stl == str) {
    int ck = d[stl][r % SQ];
    for (int i = (r % SQ) - 1; i >= l % SQ; i--) d[stl][i + 1] = d[stl][i];
    d[stl][l % SQ] = ck;
    return;
  }
  vector<int> fake;
  int tmp = r % SQ;
  tmp++;
  while (tmp) fake.push_back(d[str].front()), d[str].pop_front(), tmp--;
  int ck = fake.back();
  fake.pop_back();
  add(str, ck, -1);
  while (fake.size()) d[str].push_front(fake.back()), fake.pop_back();
  while (d[stl].size() > l % SQ)
    fake.push_back(d[stl].back()), d[stl].pop_back();
  d[stl].push_back(ck);
  add(stl, ck, 1);
  while (fake.size() > 1) d[stl].push_back(fake.back()), fake.pop_back();
  ck = fake.back();
  add(stl, ck, -1);
  for (int i = stl + 1; i < str; i++) {
    add(i, ck, 1);
    d[i].push_front(ck);
    ck = d[i].back();
    d[i].pop_back();
    add(i, ck, -1);
  }
  add(str, ck, 1);
  d[str].push_front(ck);
}
int gett(int l, int r, int k) {
  int stl = l / SQ;
  int str = r / SQ;
  int cnt = 0;
  for (int i = stl + 1; i < str; i++) cnt += get(i, k);
  for (int i = l % SQ; i < d[stl].size(); i++) {
    if (d[stl][i] == k) cnt++;
    if (stl == str && r % SQ == i) break;
  }
  if (stl < str) {
    for (int i = 0; i < (r % SQ) + 1; i++)
      if (d[str][i] == k) cnt++;
  }
  return cnt;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie();
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) {
    int ps = i / SQ;
    d[ps].push_back(a[i]);
    add(ps, a[i], 1);
  }
  int ans = 0, q;
  cin >> q;
  while (q--) {
    int t, lp, rp;
    cin >> t >> lp >> rp;
    int l = (lp + ans - 1) % n, r = (rp + ans - 1) % n;
    if (l > r) swap(l, r);
    if (t == 1) {
      shift(l, r);
    } else {
      int kp;
      cin >> kp;
      int k = (kp + ans - 1) % n + 1;
      ans = gett(l, r, k);
      cout << ans << endl;
    }
  }
  return 0;
}