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 i, j, k, l, n, m, kb;
int a[100009], szs[350], mas[350][3 * 300], cnt[350][100009];
int del(int block, int pos) {
  int val = mas[block][pos];
  cnt[block][val]--;
  for (int i = pos + 1; i <= szs[block]; i++) mas[block][i - 1] = mas[block][i];
  szs[block]--;
  return val;
}
void add(int block, int pos, int val) {
  szs[block]++;
  for (int i = szs[block]; i > pos; i--) mas[block][i] = mas[block][i - 1];
  mas[block][pos] = val;
  cnt[block][val]++;
}
pair<int, int> getnum(int num) {
  int sum = 0, s1 = num, s2 = 0;
  for (int i = 1; i <= kb; i++) {
    if (sum + szs[i] < num)
      sum += szs[i];
    else {
      return {num - sum, i};
    }
  }
  if (s2 == 0) s2 = kb;
  return {s1, s2};
}
int query(int l, int r, int val) {
  int k1 = l, k2 = r, b1, b2, sum = 0, res = 0;
  pair<int, int> s = getnum(l);
  b1 = s.second;
  k1 = s.first;
  s = getnum(r);
  b2 = s.second;
  k2 = s.first;
  for (int i = k1; i <= (b1 == b2 ? k2 : szs[b1]); i++)
    if (mas[b1][i] == val) res++;
  if (b1 != b2)
    for (int i = 1; i <= k2; i++)
      if (mas[b2][i] == val) res++;
  for (int i = b1 + 1; i <= b2 - 1; i++) res += cnt[i][val];
  return res;
}
int lastans = 0;
void form() {
  for (int i = 1; i <= n; i++) {
    mas[i / 300 + 1][++szs[i / 300 + 1]] = a[i];
    cnt[i / 300 + 1][a[i]]++;
    kb = max(kb, i / 300 + 1);
  }
}
void ins() {
  int sz = 0;
  for (int i = 1; i <= kb; i++) {
    for (int j = 1; j <= szs[i]; j++) {
      a[++sz] = mas[i][j];
      cnt[i][mas[i][j]]--;
    }
    szs[i] = 0;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  for (i = 1; i <= n; i++) cin >> a[i];
  form();
  cin >> m;
  for (i = 1; i <= m; i++) {
    if (i % 300 == 0) {
      ins();
      form();
    }
    int p;
    cin >> p >> k >> l;
    k = ((k + lastans - 1) % n) + 1;
    l = ((l + lastans - 1) % n) + 1;
    if (k > l) swap(k, l);
    if (p == 1) {
      pair<int, int> s = getnum(k);
      int b1 = s.second, k1 = s.first;
      s = getnum(l);
      int b2 = s.second, k2 = s.first;
      add(b1, k1, del(b2, k2));
    } else {
      int t;
      cin >> t;
      t = ((t + lastans - 1) % n) + 1;
      lastans = query(k, l, t);
      cout << lastans << endl;
    }
  }
  exit(0);
}