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 N = 1e5 + 50, SQ = 300, SZ = (N / SQ) + 20, INF = 0, MOD = 0;
int n, a[N], q, cnt[SZ][N];
deque<int> d[SZ];
inline void build() {
  d[0].push_back(-10);
  for (int i = 1; i <= n; i++) {
    int x = i / SQ;
    d[x].push_back(a[i]);
    cnt[x][a[i]]++;
  }
  return;
}
inline void upd(int l, int r) {
  if (l / SQ == r / SQ) {
    vector<int> tmp;
    int en = d[l / SQ].size() - 1, bg = l % SQ;
    for (int i = en; i >= bg; i--) {
      tmp.push_back(d[(l / SQ)].back());
      d[(l / SQ)].pop_back();
    }
    reverse(tmp.begin(), tmp.end());
    d[(l / SQ)].push_back(tmp[r - l]);
    for (int i = 0; i < r - l; i++) d[(l / SQ)].push_back(tmp[i]);
    for (int i = r - l + 1; i < (int)tmp.size(); i++)
      d[l / SQ].push_back(tmp[i]);
    return;
  }
  vector<int> tmp;
  int x = (r % SQ);
  for (int i = 0; i <= x; i++) {
    tmp.push_back(d[(r / SQ)].front());
    d[(r / SQ)].pop_front();
  }
  int R = tmp.back();
  cnt[(r / SQ)][R]--;
  for (int i = (int)tmp.size() - 2; i >= 0; i--) d[(r / SQ)].push_front(tmp[i]);
  for (int i = (r / SQ) - 1; i >= (l / SQ); i--) {
    int x = d[i].back();
    d[i + 1].push_front(x);
    cnt[i][x]--;
    cnt[i + 1][x]++;
    d[i].pop_back();
  }
  tmp.clear();
  int en = d[l / SQ].size() - 1, bg = l % SQ;
  for (int i = en; i >= bg; i--) {
    tmp.push_back(d[(l / SQ)].back());
    d[(l / SQ)].pop_back();
  }
  reverse(tmp.begin(), tmp.end());
  cnt[(l / SQ)][R]++;
  d[(l / SQ)].push_back(R);
  for (int i = 0; i < (int)tmp.size(); i++) d[(l / SQ)].push_back(tmp[i]);
  return;
}
int get(int l, int r, int k, int ans = 0) {
  int L = l % SQ;
  int R = r % SQ;
  if (l / SQ == r / SQ) {
    for (int i = L; i <= R; i++)
      if (d[(l / SQ)][i] == k) ans++;
    return ans;
  }
  for (int i = L; i < (int)d[l / SQ].size(); i++)
    if (d[l / SQ][i] == k) ans++;
  for (int i = (l / SQ) + 1; i < (r / SQ); i++) ans += cnt[i][k];
  for (int i = 0; i <= R; i++)
    if (d[r / SQ][i] == k) ans++;
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  build();
  cin >> q;
  int last = 0;
  for (int i = 0, l, r, k, t; i < q; i++) {
    cin >> t >> l >> r;
    l = ((l + last - 1) % n) + 1;
    r = ((r + last - 1) % n) + 1;
    if (r < l) swap(r, l);
    if (t == 1)
      upd(l, r);
    else {
      cin >> k;
      k = ((k + last - 1) % n) + 1;
      last = get(l, r, k);
      cout << last << '\n';
    }
  }
  return 0;
}