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 DIM = 120005;
const int BSIZE = 500;
deque<int> buc[BSIZE + 10];
short cnt[BSIZE][DIM];
int arr[DIM];
int _n;
inline void __(int &x, int y) { x = (x + y - 1) % _n + 1; }
pair<int, int> findPos(int p) {
  int b = 1;
  while (buc[b].size() < p) {
    p -= buc[b].size();
    ++b;
  }
  return make_pair(b, p - 1);
}
int main(void) {
  int n;
  scanf("%d", &n);
  _n = n;
  for (int i = 1; i <= n; ++i) scanf("%d", &arr[i]);
  n = (n + BSIZE - 1) / BSIZE * BSIZE;
  int q;
  scanf("%d", &q);
  int ans = 0, bnr = 0;
  for (int t = 0; t < q; ++t) {
    if (t % (BSIZE - 1) == 0) {
      int p = 0;
      for (int b = 1; b <= bnr; ++b) {
        for (int &x : buc[b]) {
          --cnt[b][x];
          arr[++p] = x;
        }
        buc[b].clear();
      }
      bnr = 1;
      for (int i = 1; i <= n; ++i) {
        if (buc[bnr].size() == BSIZE) ++bnr;
        buc[bnr].push_back(arr[i]);
        ++cnt[bnr][arr[i]];
      }
    }
    int p;
    scanf("%d", &p);
    if (p == 1) {
      int l, r;
      scanf("%d %d", &l, &r);
      __(l, ans);
      __(r, ans);
      if (l > r) swap(l, r);
      pair<int, int> pos = findPos(r);
      int nr = r - l, v = buc[pos.first][pos.second];
      while (nr and pos.second > 0) {
        swap(buc[pos.first][pos.second], buc[pos.first][pos.second - 1]);
        --pos.second;
        --nr;
      }
      if (nr > 0) {
        --cnt[pos.first][v];
        buc[pos.first].pop_front();
        --pos.first;
        while (nr > buc[pos.first].size()) {
          nr -= buc[pos.first].size();
          --pos.first;
        }
        ++cnt[pos.first][v];
        buc[pos.first].push_back(v);
        pos.second = buc[pos.first].size() - 1;
        while (nr) {
          swap(buc[pos.first][pos.second], buc[pos.first][pos.second - 1]);
          --pos.second;
          --nr;
        }
      }
    } else {
      int l, r, k;
      scanf("%d %d %d", &l, &r, &k);
      __(l, ans);
      __(r, ans);
      __(k, ans);
      if (l > r) swap(l, r);
      pair<int, int> pos1 = findPos(l), pos2 = findPos(r);
      ans = 0;
      if (pos1.first == pos2.first) {
        for (int i = pos1.second; i <= pos2.second; ++i)
          ans += (buc[pos1.first][i] == k);
      } else {
        for (int i = pos1.second; i < buc[pos1.first].size(); ++i)
          ans += (buc[pos1.first][i] == k);
        for (int i = pos1.first + 1; i <= pos2.first - 1; ++i) ans += cnt[i][k];
        for (int i = 0; i <= pos2.second; ++i) ans += (buc[pos2.first][i] == k);
      }
      printf("%d\n", ans);
    }
  }
  return 0;
}
