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;
int sq;
vector<int> v[100001];
struct node {
  int sz;
  int val;
  node* l;
  node* r;
  node* p;
  int prio;
  map<int, int> m;
  node(int va) {
    val = va;
    if (sq < v[va].size()) m[va]++;
    l = nullptr;
    r = nullptr;
    p = nullptr;
    sz = 1;
    prio = rand();
  }
};
node* ns[100001];
inline int size(const node* n) {
  if (n == nullptr) return 0;
  return n->sz;
}
void upd(node*& n) {
  if (n == nullptr) return;
  n->sz = 1 + size(n->l) + size(n->r);
  n->p = nullptr;
  if (n->l != nullptr) n->l->p = n;
  if (n->r != nullptr) n->r->p = n;
  n->m.clear();
  if (n->l != nullptr) {
    for (pair<long long, long long> p : n->l->m) {
      n->m[p.first] += p.second;
    }
  }
  if (n->r != nullptr) {
    for (pair<long long, long long> p : n->r->m) {
      n->m[p.first] += p.second;
    }
  }
  if (v[n->val].size() > sq) {
    n->m[n->val]++;
  }
}
inline void push(node*& n) {
  if (n == nullptr) return;
  n->p = nullptr;
  if (n->l != nullptr) n->l->p = n;
  if (n->r != nullptr) n->r->p = n;
}
void merge(node*& root, node* l, node* r) {
  push(l);
  push(r);
  if (l == nullptr)
    root = r;
  else if (r == nullptr)
    root = l;
  else {
    if (l->prio > r->prio) {
      merge(l->r, l->r, r);
      root = l;
    } else {
      merge(r->l, l, r->l);
      root = r;
    }
  }
  push(root);
  upd(root);
}
void split(node* n, node*& l, node*& r, const int k) {
  if (n == nullptr) {
    l = nullptr;
    r = nullptr;
    return;
  }
  push(n);
  if (size(n->l) >= k) {
    split(n->l, l, n->l, k);
    r = n;
  } else {
    split(n->r, n->r, r, k - 1 - size(n->l));
    l = n;
  }
  push(r);
  push(l);
  upd(r);
  upd(l);
}
int main() {
  cin.sync_with_stdio(false);
  cin.tie(0);
  int n;
  int q;
  srand(time(0));
  cin >> n;
  sq = (int)sqrt(n) + 1;
  node* root = nullptr;
  int w[n];
  for (int i = 0; i < n; i++) {
    cin >> w[i];
    v[w[i]].push_back(i);
  }
  for (int i = 0; i < n; i++) {
    node* n = new node(w[i]);
    ns[i] = n;
    merge(root, root, n);
  }
  cin >> q;
  int ans = 0;
  while (q-- > 0) {
    int t;
    cin >> t;
    if (t == 1) {
      int l, r;
      cin >> l >> r;
      l = ((l + ans - 1) % n) + 1;
      r = ((r + ans - 1) % n) + 1;
      if (r < l) swap(l, r);
      node* a;
      node* b;
      node* c;
      node* d;
      split(root, a, d, r);
      split(a, a, c, r - 1);
      split(a, a, b, l - 1);
      merge(root, a, c);
      merge(root, root, b);
      merge(root, root, d);
    } else if (t == 2) {
      int l, r, k;
      cin >> l >> r >> k;
      l = ((l + ans - 1) % n) + 1;
      r = ((r + ans - 1) % n) + 1;
      k = ((k + ans - 1) % n) + 1;
      if (r < l) swap(l, r);
      node* a;
      node* b;
      node* c;
      split(root, a, c, r);
      split(a, a, b, l - 1);
      if (v[k].size() <= sq) {
        int f = 0;
        for (int i : v[k]) {
          node* n = ns[i];
          while (n->p != nullptr) {
            if (n->p->p != nullptr)
              n = n->p->p;
            else
              n = n->p;
          }
          if (n == b) f++;
        }
        cout << f << '\n';
        ans = f;
      } else {
        int f = b->m[k];
        cout << f << '\n';
        ans = f;
      }
      merge(root, a, b);
      merge(root, root, c);
    }
  }
}