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 N = 100005;
const int block = 318;
const int C = 318;
vector<int> a;
int n, q, cblock;
int num[block + 5][N], sz[block + 5], wsz[block + 5];
int val[block + 5][block + C + 5], wval[block + 5][block + C + 5];
void build() {
  for (int i = 1; i <= n; i += block) {
    ++cblock;
    int r = min(n, i + block - 1);
    for (int j = i; j <= r; ++j) {
      val[cblock][++sz[cblock]] = a[j];
      wval[cblock][++wsz[cblock]] = a[j];
      ++num[cblock][a[j]];
    }
  }
}
void rebuild() {
  int id = 0;
  for (int i = 1; i <= cblock; ++i) {
    for (int j = 1; j <= sz[i]; ++j) a[++id] = val[i][j];
    sz[i] = 0;
    for (int j = 1; j <= wsz[i]; ++j) num[i][wval[i][j]] = 0;
    wsz[i] = 0;
  }
  cblock = 0;
  build();
}
int del(int id) {
  int sum = 0;
  for (int i = 1; i <= cblock; ++i) {
    sum += sz[i];
    if (sum >= id) {
      int pos = sz[i] + id - sum;
      int res = val[i][pos];
      for (int j = pos; j < sz[i]; ++j) val[i][j] = val[i][j + 1];
      --sz[i];
      --num[i][res];
      return res;
    }
  }
  return 0;
}
void add(int id, int x) {
  int sum = 0;
  for (int i = 1; i <= cblock; ++i) {
    sum += sz[i];
    if (sum >= id) {
      int pos = sz[i] + id - sum;
      ++sz[i];
      for (int j = sz[i]; j > pos; --j) val[i][j] = val[i][j - 1];
      val[i][pos] = x;
      wval[i][++wsz[i]] = x;
      ++num[i][x];
      break;
    }
  }
}
void shift(int l, int r) {
  if (l == r) return;
  int x = del(r);
  add(l, x);
}
int cal(int id, int x) {
  int sum = 0, res = 0;
  for (int i = 1; i <= cblock; ++i) {
    sum += sz[i];
    if (sum > id) {
      int pos = sz[i] + id - sum;
      for (int j = 1; j <= pos; ++j)
        if (val[i][j] == x) ++res;
      break;
    }
    res += num[i][x];
  }
  return res;
}
int cal(int l, int r, int x) { return cal(r, x) - cal(l - 1, x); }
void init() {
  cin >> n;
  a.resize(n + 1);
  for (int i = 1; i <= n; ++i) cin >> a[i];
  build();
  cin >> q;
}
void solve() {
  int last = 0, l, r, k, cnt = 0;
  while (q--) {
    ++cnt;
    int op;
    cin >> op;
    if (op == 1) {
      cin >> l >> r;
      l = ((l + last - 1 + n) % n) + 1;
      r = ((r + last - 1 + n) % n) + 1;
      if (l > r) swap(l, r);
      shift(l, r);
    } else {
      cin >> l >> r >> k;
      l = ((l + last - 1 + n) % n) + 1;
      r = ((r + last - 1 + n) % n) + 1;
      k = ((k + last - 1 + n) % n) + 1;
      if (l > r) swap(l, r);
      last = cal(l, r, k);
      cout << last << '\n';
    }
    if (cnt % (C - 5) == 0) rebuild();
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  init();
  solve();
}