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>
const int MAXN = 1e5 + 19, MAXB = 400;
int n, b, q, belong[MAXN];
int lastans;
std::deque<int> a[MAXB];
int cnt[MAXB][MAXN];
inline void move(int node) {
  ++cnt[node + 1][a[node].back()];
  --cnt[node][a[node].back()];
  a[node + 1].push_front(a[node].back());
  a[node].pop_back();
}
void modify(int l, int r) {
  int vr = a[belong[r]][r - (belong[r] - 1) * b - 1];
  if (belong[l] == belong[r]) {
    for (int i = r - (belong[l] - 1) * b - 1; i >= l - (belong[l] - 1) * b; --i)
      a[belong[l]][i] = a[belong[l]][i - 1];
    a[belong[l]][l - (belong[l] - 1) * b - 1] = vr;
    return;
  }
  for (int i = belong[l]; i < belong[r]; ++i) move(i);
  ++cnt[belong[l]][vr];
  --cnt[belong[r]][vr];
  std::vector<int> tmp;
  int pos = l - (belong[l] - 1) * b - 1;
  for (int i = 0; i < a[belong[l]].size(); ++i) {
    if (i == pos) tmp.push_back(vr);
    tmp.push_back(a[belong[l]][i]);
  }
  if (pos == a[belong[l]].size()) tmp.push_back(vr);
  a[belong[l]] = std::deque<int>(tmp.begin(), tmp.end());
  tmp.clear();
  pos = r - (belong[r] - 1) * b;
  for (int i = 0; i < a[belong[r]].size(); ++i)
    if (i != pos) tmp.push_back(a[belong[r]][i]);
  a[belong[r]] = std::deque<int>(tmp.begin(), tmp.end());
}
int query(int l, int r, int k) {
  int ans = 0;
  if (belong[l] == belong[r]) {
    for (int i = l - (belong[l] - 1) * b - 1; i <= r - (belong[l] - 1) * b - 1;
         ++i)
      ans += (a[belong[l]][i] == k);
    return ans;
  }
  for (int i = belong[l] + 1; i < belong[r]; ++i) ans += cnt[i][k];
  for (int i = l - (belong[l] - 1) * b - 1; i < a[belong[l]].size(); ++i)
    ans += (a[belong[l]][i] == k);
  for (int i = 0; i <= r - (belong[r] - 1) * b - 1; ++i)
    ans += (a[belong[r]][i] == k);
  return ans;
}
int main() {
  std::scanf("%d", &n);
  b = std::sqrt(n + 1);
  for (int i = 1; i <= n; ++i) belong[i] = (i - 1) / b + 1;
  for (int i = 1; i <= n; ++i) {
    int tmp;
    std::scanf("%d", &tmp);
    a[belong[i]].push_back(tmp);
    ++cnt[belong[i]][tmp];
  }
  std::scanf("%d", &q);
  while (q--) {
    int opt, l, r;
    std::cin >> opt >> l >> r;
    l = (l + lastans - 1) % n + 1;
    r = (r + lastans - 1) % n + 1;
    if (l > r) std::swap(l, r);
    if (opt == 1)
      modify(l, r);
    else if (opt == 2) {
      int k;
      std::scanf("%d", &k);
      k = (k + lastans - 1) % n + 1;
      std::printf("%d\n", lastans = query(l, r, k));
    }
  }
  return 0;
}