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", "unroll-loops")
using namespace std;
const long long inf = numeric_limits<long long>::max() / 2;
const long double eps = 1e-9;
const long double pi = acos(-1);
struct sqrtd {
  vector<long long> a;
  vector<vector<long long> > t;
  vector<vector<long long> > cnt;
  vector<long long> bs;
  long long SQRT = 6237;
  sqrtd() {}
  sqrtd(vector<long long> as) {
    a = as;
    build();
  }
  void build() {
    long long n = (long long)((a).size());
    bs.clear();
    t.assign(n / SQRT + 1, {});
    bs.assign(n / SQRT + 1, 0);
    cnt.assign(n / SQRT + 1, {});
    for (long long i = 0; i < n; i++) {
      t[i / SQRT].push_back(a[i]);
      bs[i / SQRT] += a[i];
      if (cnt[i / SQRT].size() <= a[i]) {
        cnt[i / SQRT].resize(a[i] + 1);
      }
      ++cnt[i / SQRT][a[i]];
    }
  }
  void rebuild() {
    a.clear();
    for (long long i = 0; i < (long long)((t).size()); i++) {
      for (long long j = 0; j < (long long)((t[i]).size()); j++) {
        a.push_back(t[i][j]);
      }
    }
    build();
  }
  void print() {
    for (long long i = 0; i < (long long)((t).size()); i++) {
      for (long long j = 0; j < (long long)((t[i]).size()); j++) {
        cout << t[i][j] << " ";
      }
    }
    cout << endl;
  }
  long long get(long long k, long long L, long long R) {
    long long ans = 0;
    long long cl = 0, cr = 0;
    for (long long i = 0; i < (long long)((t).size()); i++) {
      if (t[i].empty()) {
        break;
      }
      cr = cl + (long long)((t[i]).size()) - 1;
      if (L <= cl && cr <= R) {
        if (cnt[i].size() > k) {
          ans += cnt[i][k];
        }
      } else {
        for (long long j = max(cl, L); j <= min(cr, R); j++) {
          if (t[i][j - cl] == k) {
            ++ans;
          }
        }
      }
      cl += (long long)((t[i]).size());
    }
    return ans;
  }
  void rotate(long long L, long long R) {
    long long cl = 0, cr = 0;
    pair<long long, long long> lpos = {-1, -1}, rpos = {-1, -1};
    for (int i = 0; i < (long long)((t).size()); i++) {
      cr = cl + (long long)((t[i]).size()) - 1;
      if (cl <= L && L <= cr) {
        lpos = {i, L - cl};
      }
      if (cl <= R && R <= cr) {
        rpos = {i, R - cl};
      }
      cl += (long long)((t[i]).size());
    }
    long long rg = t[rpos.first][rpos.second];
    t[rpos.first].erase(t[rpos.first].begin() + rpos.second);
    t[lpos.first].insert(t[lpos.first].begin() + lpos.second, rg);
    --cnt[rpos.first][rg];
    if ((long long)((cnt[lpos.first]).size()) <= rg) {
      cnt[lpos.first].resize(rg + 1);
    }
    ++cnt[lpos.first][rg];
  }
};
long long N, Q;
vector<long long> a;
sqrtd s;
void solve() {
  long long prev = 0;
  s = sqrtd(a);
  for (long long i = 0; i < Q; i++) {
    if (i % s.SQRT == 0) {
      s.rebuild();
    }
    long long op, L, R;
    cin >> op >> L >> R;
    L = ((L + prev - 1) % N) + 1;
    R = ((R + prev - 1) % N) + 1;
    --L, --R;
    if (R < L) {
      swap(L, R);
    }
    if (op == 1) {
      s.rotate(L, R);
    } else {
      long long k;
      cin >> k;
      k = ((k + prev - 1) % N) + 1;
      prev = s.get(k, L, R);
      cout << prev << "\n";
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> N;
  a.resize(N);
  for (long long i = 0; i < N; i++) {
    cin >> a[i];
  }
  cin >> Q;
  solve();
  return 0;
}