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>
constexpr size_t R = 256;
int main() {
  size_t N;
  std::cin >> N;
  std::vector<std::deque<size_t>> V;
  std::vector<std::vector<size_t>> C;
  for (size_t i{}; i < N; ++i) {
    size_t a;
    std::cin >> a;
    --a;
    if (!V.size() || V.back().size() == R) {
      V.emplace_back();
      C.emplace_back(N, 0);
    }
    V.back().push_back(a);
    C.back()[a] += 1;
  }
  size_t Q;
  std::cin >> Q;
  size_t lastans{};
  for (char c; std::cin >> c;) switch (c) {
      case '1': {
        size_t l, r;
        std::cin >> l >> r;
        l = (l + lastans - 1) % N;
        r = (r + lastans - 1) % N;
        if (l > r) std::swap(l, r);
        auto k = V[r / R][r % R];
        V[r / R].erase(V[r / R].begin() + r % R);
        C[r / R][k] -= 1;
        V[l / R].insert(V[l / R].begin() + l % R, k);
        C[l / R][k] += 1;
        for (size_t i{l / R}; i + 1 <= r / R; ++i) {
          auto v = V[i].back();
          V[i].pop_back();
          C[i][v] -= 1;
          V[i + 1].push_front(v);
          C[i + 1][v] += 1;
        }
        break;
      }
      case '2': {
        size_t l, r, k;
        std::cin >> l >> r >> k;
        l = (l + lastans - 1) % N;
        r = (r + lastans - 1) % N;
        k = (k + lastans - 1) % N;
        if (l > r) std::swap(l, r);
        ++r;
        lastans = 0;
        for (size_t i{l}; i < r;)
          if (i % R == 0 && i + R <= r) {
            lastans += C[i / R][k];
            i += R;
          } else {
            lastans += V[i / R][i % R] == k;
            i += 1;
          }
        std::cout << lastans << '\n';
        break;
      }
      default:
        __builtin_unreachable();
    }
}