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 MAX = 1e5 + 9, MAX2 = 400;
int a[MAX], fre[MAX2][MAX], n, seg, segNo;
vector<int> block[MAX2];
void build() {
  for (int i = 0; i < n; i++) {
    block[i / seg].push_back(a[i]);
    fre[i / seg][a[i]]++;
  }
}
pair<int, int> findBlock(int ind) {
  int sum = 0;
  for (int i = 0; i < segNo; i++) {
    sum += block[i].size();
    if (sum > ind) {
      sum -= block[i].size();
      return {i, sum};
    }
  }
  return {segNo - 1, sum - block[segNo - 1].size()};
}
void insert(int ind, int val) {
  pair<int, int> temp = findBlock(ind);
  int bi = temp.first, pos = ind - temp.second;
  block[bi].insert(block[bi].begin() + pos, val);
  fre[bi][val]++;
}
int erase(int ind) {
  pair<int, int> temp = findBlock(ind);
  int bi = temp.first, pos = ind - temp.second;
  int val = block[bi][pos];
  block[bi].erase(block[bi].begin() + pos);
  fre[bi][val]--;
  return val;
}
void circShift(int l, int r) {
  int val = erase(r);
  insert(l, val);
}
void reBuild() {
  for (int i = 0; i < segNo; i++) {
    if ((int)block[i].size() > seg) {
      auto start = block[i].begin() + seg;
      for (auto j = start; j != block[i].end(); j++)
        fre[i][*j]--, fre[i + 1][*j]++;
      block[i + 1].insert(block[i + 1].begin(), start, block[i].end());
      block[i].erase(start, block[i].end());
    }
  }
}
int query(int l, int r, int k) {
  pair<int, int> start = findBlock(l), stop = findBlock(r);
  int sum = 0;
  if (start.first == stop.first) {
    for (int i = l - start.second; i <= r - stop.second; i++) {
      if (block[start.first][i] == k) sum++;
    }
  } else {
    for (int i = l - start.second; i < (int)block[start.first].size(); i++) {
      if (block[start.first][i] == k) sum++;
    }
    for (int i = start.first + 1; i < stop.first; i++) {
      sum += fre[i][k];
    }
    for (int i = 0; i <= r - stop.second; i++) {
      if (block[stop.first][i] == k) sum++;
    }
  }
  return sum;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  seg = ceil(sqrt(n)), segNo = (n - 1) / seg + 1;
  build();
  int q, lastans = 0;
  cin >> q;
  for (int i = 0, count = 0; i < q; i++) {
    if (count > seg) reBuild(), count = 0;
    int t;
    long long l, r, k;
    cin >> t;
    if (t == 1) {
      cin >> l >> r;
      l = (l + lastans - 1) % n;
      r = (r + lastans - 1) % n;
      if (l > r) swap(l, r);
      circShift(l, r);
      count++;
    } else {
      cin >> l >> r >> k;
      l = (l + lastans - 1) % n;
      r = (r + lastans - 1) % n;
      k = (k + lastans - 1) % n + 1;
      if (l > r) swap(l, r);
      lastans = query(l, r, k);
      cout << lastans << '\n';
    }
  }
  return 0;
}