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 MOD = 998244353;
const int oo = 1e9;
const int MAXN = 2e5 + 5, MAXBLK = 320;
int a[MAXN];
short cnt[MAXBLK][MAXN];
vector<int> block[MAXBLK];
int n, sz_bck;
void build() {
  int cur_blk = -1;
  for (int i = 0; i < n; ++i) {
    if (i % sz_bck == 0) cur_blk++;
    block[cur_blk].push_back(a[i]);
    cnt[cur_blk][a[i]]++;
  }
}
void rebuild() {
  int cur_blk = 0;
  for (int i = 0; i < n;) {
    for (int j = 0; j < (int)(block[cur_blk]).size(); ++i, ++j) {
      a[i] = block[cur_blk][j];
      cnt[cur_blk][a[i]]--;
    }
    block[cur_blk].clear();
    cur_blk++;
  }
  build();
}
void shift(int l, int r) {
  int cur_blk = 0;
  while (r >= (int)(block[cur_blk]).size()) {
    r -= (int)(block[cur_blk]).size();
    ++cur_blk;
  }
  int elem = block[cur_blk][r];
  block[cur_blk].erase(begin(block[cur_blk]) + r);
  cnt[cur_blk][elem]--;
  cur_blk = 0;
  while (l >= (int)(block[cur_blk]).size()) {
    l -= (int)(block[cur_blk]).size();
    ++cur_blk;
  }
  block[cur_blk].insert(begin(block[cur_blk]) + l, elem);
  cnt[cur_blk][elem]++;
}
int query(int l, int r, int k) {
  int cur_blk = 0;
  int range = r - l + 1;
  while (l >= (int)(block[cur_blk]).size()) {
    l -= (int)(block[cur_blk]).size();
    ++cur_blk;
  }
  int ans = 0;
  while (l < (int)(block[cur_blk]).size() && range) {
    ans += block[cur_blk][l] == k;
    ++l, --range;
  }
  ++cur_blk;
  while ((int)(block[cur_blk]).size() <= range && range) {
    ans += cnt[cur_blk][k];
    range -= (int)(block[cur_blk]).size();
    ++cur_blk;
  }
  while (range) {
    ans += block[cur_blk][range - 1] == k;
    --range;
  }
  return ans;
}
void debug_blocks() {
  int cur_blk = 0;
  for (int i = 0; i < n;) {
    cerr << "[ ";
    for (int j = 0; j < (int)(block[cur_blk]).size(); ++i, ++j) {
      cerr << block[cur_blk][j] << " ";
    }
    cerr << "] ";
    ++cur_blk;
  }
  cerr << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  sz_bck = sqrt(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  build();
  int q, type1 = 0, ans = 0;
  cin >> q;
  while (q--) {
    int op, l, r, k;
    cin >> op >> l >> r;
    l = (l + ans - 1) % n + 1;
    r = (r + ans - 1) % n + 1;
    --l, --r;
    if (l > r) swap(l, r);
    if (op == 1) {
      if (l != r) shift(l, r);
      type1++;
    } else {
      cin >> k;
      k = (k + ans - 1) % n + 1;
      ans = query(l, r, k);
      cout << ans << "\n";
    }
    if (type1 == sz_bck) {
      rebuild();
      type1 = 0;
    }
  }
  return 0;
}
