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 SZ = 1230;
int N, Q, lastans;
struct block {
  int L, R;
  int freq[100001];
  list<int> data;
  int update(int l, int r, int x) {
    int p = max(l, L), q = min(r, R);
    int ret = 0;
    if (L <= l && r <= R) {
      auto it = data.begin();
      advance(it, r - L);
      int v = *it;
      data.erase(it);
      it = data.begin();
      advance(it, l - L);
      data.insert(it, v);
    } else if (p == L && q == R) {
      freq[x]++;
      data.push_front(x);
      ret = data.back();
      data.pop_back();
      freq[ret]--;
    } else if (q == R) {
      freq[x]++;
      auto it = data.begin();
      advance(it, l - L);
      data.insert(it, x);
      ret = data.back();
      data.pop_back();
      freq[ret]--;
    } else if (p == L) {
      auto it = data.begin();
      advance(it, r - L);
      freq[*it]--;
      data.erase(it);
      freq[x]++;
      data.push_front(x);
    }
    return ret;
  }
  int query(int l, int r, int x) {
    int p = max(l, L), q = min(r, R);
    if (p == L && q == R) return freq[x];
    int ret = 0;
    auto it = data.begin();
    advance(it, p - L);
    for (int i = p; i <= q; i++, ++it) ret += (*it) == x;
    return ret;
  }
} blocks[100000 / SZ + 5];
int main() {
  scanf("%d", &N);
  int a, b, c, d;
  int cur = 0, last = -1;
  for (int i = 1; i <= N; i++) {
    cur = i / SZ;
    if (cur != last) {
      if (last != -1) blocks[last].R = i - 1;
      blocks[cur].L = i;
      last = cur;
    }
    scanf("%d", &a);
    blocks[cur].data.push_back(a);
    blocks[cur].freq[a]++;
  }
  blocks[cur].R = N;
  scanf("%d", &Q);
  lastans = 0;
  for (int i = 0; i < Q; i++) {
    scanf("%d%d%d", &a, &b, &c);
    b = (b + lastans - 1) % N + 1;
    c = (c + lastans - 1) % N + 1;
    if (b > c) swap(b, c);
    int p = b / SZ, q = c / SZ;
    if (a == 1) {
      if (b == c) continue;
      auto it = blocks[q].data.begin();
      advance(it, c - blocks[q].L);
      int last = *it;
      for (int i = p; i <= q; i++) last = blocks[i].update(b, c, last);
    } else {
      scanf("%d", &d);
      d = (d + lastans - 1) % N + 1;
      int ans = 0;
      for (int i = p; i <= q; i++) ans += blocks[i].query(b, c, d);
      printf("%d\n", ans);
      lastans = ans;
    }
  }
  return 0;
}