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;
using LL = int64_t;
using pii = pair<int, int>;
const double EPS = 1e-9;
const double PI = 2 * asin(1);
const int INF = 0x3f3f3f3f;
const LL LINF = 0x3f3f3f3f3f3f3f3f;
const int MX = 100000 + 10;
const int NM = 318;
int g_n, a[MX], cnt[NM][MX];
deque<deque<int>> ddi;
void pretreat() {}
bool input() {
  cin >> g_n;
  if (cin.eof()) return false;
  memset(cnt, 0, sizeof(cnt));
  ddi.clear(), ddi.resize(NM + 1);
  for (int i = 0, t; i < g_n; ++i) {
    scanf(" %d", &t);
    ++cnt[i / NM][t];
    ddi[i / NM].push_back(t);
  }
  return true;
}
void solve() {
  int ans = 0, q;
  cin >> q;
  while (q--) {
    int t, u, v, k;
    scanf(" %d %d %d", &t, &u, &v);
    u = (u + ans - 1) % g_n;
    v = (v + ans - 1) % g_n;
    if (u > v) swap(u, v);
    int block_u = u / NM, pos_u = u % NM;
    int block_v = v / NM, pos_v = v % NM;
    if (t == 1) {
      k = ddi[block_v][pos_v];
      ddi[block_v].erase(ddi[block_v].begin() + pos_v);
      ddi[block_u].insert(ddi[block_u].begin() + pos_u, k);
      ++cnt[block_u][k], --cnt[block_v][k];
      for (int i = block_u; i < block_v; ++i) {
        k = ddi[i].back();
        ddi[i].pop_back();
        ddi[i + 1].push_front(k);
        ++cnt[i + 1][k], --cnt[i][k];
      }
    } else if (t == 2) {
      scanf(" %d", &k);
      k = (k + ans - 1) % g_n + 1;
      ans = 0;
      if (block_u == block_v) {
        for (int i = pos_u; i <= pos_v; ++i) {
          if (ddi[block_u][i] == k) ++ans;
        }
      } else {
        for (int i = pos_u; i < NM; ++i) {
          if (ddi[block_u][i] == k) ++ans;
        }
        for (int i = block_u + 1; i < block_v; ++i) {
          ans += cnt[i][k];
        }
        for (int i = 0; i <= pos_v; ++i) {
          if (ddi[block_v][i] == k) ++ans;
        }
      }
      printf("%d\n", ans);
    }
  }
}
int main() {
  pretreat();
  while (input()) {
    solve();
  }
  return 0;
}