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 maxn = 1e5 + 233;
const int block_size = 400;
vector<int> cnt[block_size];
int pl[maxn], pr[maxn];
int bel[maxn];
deque<int> q[block_size];
int n, m;
void update(int o, int l, int r) {
  deque<int> t;
  for (int i = 0; i < l; i++) {
    t.push_back(q[o][i]);
  }
  t.push_back(q[o][r]);
  for (int i = l; i < r; i++) {
    t.push_back(q[o][i]);
  }
  for (int i = r + 1; i < q[o].size(); i++) {
    t.push_back(q[o][i]);
  }
  q[o] = t;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    bel[i] = i / block_size + 1;
  }
  for (int i = 1; i <= bel[n]; i++) {
    cnt[i].resize(n + 1, 0);
  }
  for (int i = 1; i <= n; ++i) {
    int x;
    cin >> x;
    q[bel[i]].push_back(x);
    cnt[bel[i]][x]++;
  }
  for (int i = 1; i <= n; ++i) {
    pr[bel[i]] = i;
  }
  for (int i = n; i; --i) {
    pl[bel[i]] = i;
  }
  cin >> m;
  int last = 0;
  while (m--) {
    int op, l, r, x;
    cin >> op >> l >> r;
    l = (l + last - 1) % n + 1;
    r = (r + last - 1) % n + 1;
    if (l > r) swap(l, r);
    if (op == 1) {
      if (bel[l] == bel[r]) {
        update(bel[l], l - pl[bel[l]], r - pl[bel[l]]);
        continue;
      }
      int p = q[bel[r]][r - pl[bel[r]]];
      update(bel[r], 0, r - pl[bel[r]]);
      cnt[bel[r]][p]--;
      q[bel[r]][0] = q[bel[r] - 1].back();
      cnt[bel[r]][q[bel[r]][0]]++;
      for (int i = bel[r] - 1; i > bel[l]; i--) {
        cnt[i][q[i].back()]--;
        q[i].pop_back();
        q[i].push_front(q[i - 1].back());
        cnt[i][q[i][0]]++;
      }
      cnt[bel[l]][q[bel[l]].back()]--;
      update(bel[l], l - pl[bel[l]], pr[bel[l]] - pl[bel[l]]);
      q[bel[l]][l - pl[bel[l]]] = p;
      cnt[bel[l]][p]++;
    } else {
      cin >> x;
      x = (x + last - 1) % n + 1;
      if (bel[l] == bel[r]) {
        int res = 0;
        for (int i = l - pl[bel[l]]; i <= r - pl[bel[l]]; i++) {
          res += q[bel[l]][i] == x;
        }
        last = res;
        cout << last << "\n";
        continue;
      }
      int res = 0;
      for (int i = l - pl[bel[l]]; i <= pr[bel[l]] - pl[bel[l]]; i++) {
        res += q[bel[l]][i] == x;
      }
      for (int i = 0; i <= r - pl[bel[r]]; i++) {
        res += q[bel[r]][i] == x;
      }
      for (int i = bel[l] + 1; i < bel[r]; i++) {
        res += cnt[i][x];
      }
      last = res;
      cout << last << "\n";
    }
  }
  return 0;
}