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 = 1e5 + 5, blocksize = 350;
int n, a[N], cnt[400][N], q, type;
deque<int> dq[400];
void shift(int L, int R) {
  int blockL = L / blocksize, blockR = R / blocksize;
  int posR = R % blocksize, posL = L % blocksize;
  deque<int>::iterator it;
  it = (dq[blockR].begin() + posR);
  int val = *it;
  dq[blockR].erase(it);
  cnt[blockR][val]--;
  it = (dq[blockL].begin() + posL);
  dq[blockL].insert(it, val);
  cnt[blockL][val]++;
  int curBlock = blockL;
  while (curBlock < blockR) {
    int Back = dq[curBlock].back();
    dq[curBlock + 1].push_front(Back);
    ++cnt[curBlock + 1][Back];
    dq[curBlock].pop_back();
    --cnt[curBlock][Back];
    ++curBlock;
  }
}
int solve(int L, int R, int K) {
  int blockL = L / blocksize, blockR = R / blocksize;
  int posR = R % blocksize, posL = L % blocksize, res = 0;
  for (int i = blockL + 1; i <= blockR - 1; ++i) res += cnt[i][K];
  if (blockL != blockR) {
    for (auto it = (dq[blockL].begin() + posL); it != dq[blockL].end(); ++it)
      res += ((*it) == K);
    for (auto it = dq[blockR].begin(); it != (dq[blockR].begin() + posR + 1);
         ++it)
      res += ((*it) == K);
  } else
    for (auto it = (dq[blockL].begin() + posL);
         it != (dq[blockR].begin() + posR + 1); ++it)
      res += ((*it) == K);
  return res;
}
int main() {
  int lastans = 0, li, ri, ki, k, l, r;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    ++cnt[i / blocksize][a[i]];
    dq[i / blocksize].push_back(a[i]);
  }
  cin >> q;
  while (q--) {
    cin >> type >> li >> ri;
    l = (li + lastans - 1) % n;
    r = (ri + lastans - 1) % n;
    if (l > r) swap(l, r);
    if (type == 1) {
      shift(l, r);
    } else {
      cin >> ki;
      k = (ki + lastans - 1) % n + 1;
      lastans = solve(l, r, k);
      cout << lastans << '\n';
    }
  }
}