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>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
using namespace std;
const long long nmax = 100005;
long long n, q, max_block, sz, val, a[nmax], bucket[320][nmax];
vector<long long> bucket_el[320];
void build() {
  for (long long i = 1; i <= n; i++) {
    bucket[i / sz][a[i]]++;
    bucket_el[i / sz].push_back(a[i]);
  }
}
void rem(long long pos) {
  long long curr = 0;
  while (true) {
    if (pos - (long long)bucket_el[curr].size() > 0) {
      pos -= (long long)bucket_el[curr].size();
      ++curr;
    } else {
      val = bucket_el[curr][pos - 1];
      vector<long long> arr;
      bucket[curr][bucket_el[curr][pos - 1]]--;
      for (long long i = 0; i < pos - 1; i++) {
        arr.push_back(bucket_el[curr][i]);
      }
      for (long long i = pos; i < (long long)bucket_el[curr].size(); i++) {
        arr.push_back(bucket_el[curr][i]);
      }
      bucket_el[curr] = arr;
      break;
    }
  }
}
void add(long long pos, long long val) {
  long long curr = 0;
  while (true) {
    if (pos - (long long)bucket_el[curr].size() > 1) {
      pos -= (long long)bucket_el[curr].size();
      ++curr;
    } else {
      vector<long long> arr;
      bucket[curr][val]++;
      for (long long i = 0; i < pos - 1; i++) {
        arr.push_back(bucket_el[curr][i]);
      }
      arr.push_back(val);
      for (long long i = pos - 1; i < (long long)bucket_el[curr].size(); i++) {
        arr.push_back(bucket_el[curr][i]);
      }
      bucket_el[curr] = arr;
      break;
    }
  }
}
long long query(long long pos, long long crt) {
  long long curr = 0;
  long long ans = 0;
  while (true) {
    if (pos - (long long)bucket_el[curr].size() > 0) {
      pos -= (long long)bucket_el[curr].size();
      ans += bucket[curr][crt];
      ++curr;
    } else {
      for (long long i = 0; i <= pos - 1; i++) {
        if (bucket_el[curr][i] == crt) ans++;
      }
      break;
    }
  }
  return ans;
}
void rebuild() {
  vector<long long> arr;
  for (long long i = 0; i <= max_block; i++) {
    for (auto it : bucket_el[i]) {
      arr.push_back(it);
      bucket[i][it] = 0;
    }
    bucket_el[i].clear();
  }
  for (long long i = 0; i < (long long)arr.size(); i++) {
    bucket[(i + 1) / sz][arr[i]]++;
    bucket_el[(i + 1) / sz].push_back(arr[i]);
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cerr.tie(0);
  cout.tie(0);
  srand(chrono::steady_clock::now().time_since_epoch().count());
  cin >> n;
  sz = sqrt(n) + 1;
  max_block = n / sz;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
  }
  build();
  long long last = 0;
  cin >> q;
  for (long long i = 1; i <= q; i++) {
    long long type;
    cin >> type;
    if (type == 1) {
      long long l, r;
      cin >> l >> r;
      l = (l + last - 1) % n + 1;
      r = (r + last - 1) % n + 1;
      if (r < l) swap(l, r);
      val = 0;
      rem(r);
      add(l, val);
    } else {
      long long l, r, k;
      cin >> l >> r >> k;
      l = (l + last - 1) % n + 1;
      r = (r + last - 1) % n + 1;
      k = (k + last - 1) % n + 1;
      if (r < l) swap(l, r);
      last = query(r, k) - query(l - 1, k);
      cout << last << '\n';
    }
    if (i % sz == 0) {
      rebuild();
    }
  }
}