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;
int const lim = 1e5 + 3;
int const Msq = 320;
int inp[lim + 3], mx_con;
vector<int> cur[320];
int sq, cnt[Msq][lim + 3], n;
void build() {
  vector<int> tem;
  for (int i = 0; i <= sq; i++) {
    for (int j = 0; j < cur[i].size(); j++)
      tem.push_back(cur[i][j]), cnt[i][cur[i][j]]--;
    cur[i].clear();
  }
  for (int i = 0; i < tem.size(); i++)
    cur[i / sq].push_back(tem[i]), cnt[i / sq][tem[i]]++;
}
void update(int l, int r) {
  int con_a = 0, con_b = 0, pre_a = 0, pre_b = 0;
  while (pre_a + cur[con_a].size() - 1 < l) pre_a += cur[con_a].size(), con_a++;
  con_b = con_a, pre_b = pre_a;
  while (pre_b + cur[con_b].size() - 1 < r) pre_b += cur[con_b].size(), con_b++;
  int val = cur[con_b][r - pre_b];
  int ind_b = r - pre_b;
  cur[con_b].erase(cur[con_b].begin() + ind_b);
  cur[con_a].insert(cur[con_a].begin() + (l - pre_a), val);
  cnt[con_a][val]++;
  cnt[con_b][val]--;
}
int query(int l, int r, int x) {
  int ret = 0;
  int con_a = 0, con_b = 0, pre_a = 0, pre_b = 0;
  while (pre_a + cur[con_a].size() - 1 < l) pre_a += cur[con_a].size(), con_a++;
  con_b = con_a, pre_b = pre_a;
  while (pre_b + cur[con_b].size() - 1 < r) pre_b += cur[con_b].size(), con_b++;
  int ind_l = l - pre_a, ind_r = r - pre_b;
  if (con_a == con_b) {
    for (int i = ind_l; i <= ind_r; i++)
      if (cur[con_a][i] == x) ret++;
    return ret;
  }
  for (int i = ind_l; i < cur[con_a].size(); i++)
    if (cur[con_a][i] == x) ret++;
  for (int i = 0; i <= ind_r; i++)
    if (cur[con_b][i] == x) ret++;
  for (int i = con_a + 1; i < con_b; i++) ret += cnt[i][x];
  return ret;
}
int main() {
  cin >> n;
  sq = sqrt(n) + 1;
  for (int i = 0; i < n; i++)
    scanf("%d", &inp[i]), cur[i / sq].push_back(inp[i]), cnt[i / sq][inp[i]]++;
  int last_ans = 0, m;
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    if (i && i % sq == 0) build();
    int t;
    scanf("%d", &t);
    int b, e;
    scanf("%d%d", &b, &e);
    b = (b + last_ans - 1 + n) % n;
    e = (e + last_ans - 1 + n) % n;
    if (b > e) swap(e, b);
    if (t == 1)
      update(b, e);
    else {
      int x;
      scanf("%d", &x);
      x = (x + last_ans - 1 + n) % n;
      x++;
      last_ans = query(b, e, x);
      printf("%d\n", last_ans);
    }
  }
}
