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 inf = 1e5 + 9, sqr = 322;
int n, q, a[inf];
struct B {
  int st, en, done = 0;
  int cnt[inf];
  deque<int> dq;
} blocks[sqr + 44];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    blocks[i / sqr].en = i;
    if (!blocks[i / sqr].done) blocks[i / sqr].st = i, blocks[i / sqr].done = 1;
    blocks[i / sqr].dq.push_back(a[i]);
    blocks[i / sqr].cnt[a[i]]++;
  }
  scanf("%d", &q);
  int last = 0;
  while (q--) {
    int type, l, r, k, ans = 0;
    scanf("%d%d%d", &type, &l, &r);
    l = (l + last - 1) % n + 1;
    r = (r + last - 1) % n + 1;
    if (l > r) swap(l, r);
    l--, r--;
    if (type == 2) {
      scanf("%d", &k);
      k = (k + last - 1) % n + 1;
      if (l / sqr == r / sqr) {
        while (l <= r)
          ans += (blocks[l / sqr].dq[l - blocks[l / sqr].st] == k), l++;
        printf("%d\n", ans);
        last = ans;
        continue;
      }
      while (l % sqr != 0)
        ans += (blocks[l / sqr].dq[l - blocks[l / sqr].st] == k), l++;
      while (r % sqr != sqr - 1)
        ans += (blocks[r / sqr].dq[r - blocks[r / sqr].st] == k), r--;
      while (l / sqr <= r / sqr) ans += blocks[l / sqr].cnt[k], l += sqr;
      last = ans;
      printf("%d\n", ans);
    } else {
      if (l / sqr == r / sqr) {
        vector<int> v;
        int val = blocks[l / sqr].dq[r - blocks[l / sqr].st];
        int idx = blocks[l / sqr].st;
        for (auto o : blocks[l / sqr].dq) {
          if (idx == l) v.push_back(val);
          if (idx != r) v.push_back(o);
          idx++;
        }
        blocks[l / sqr].dq.clear();
        for (auto o : v) blocks[l / sqr].dq.push_back(o);
        continue;
      }
      vector<int> en, st;
      int val, idx;
      idx = blocks[r / sqr].st;
      val = blocks[r / sqr].dq[r - blocks[r / sqr].st];
      for (auto o : blocks[r / sqr].dq) {
        if (idx != r) en.push_back(o);
        idx++;
      }
      idx = blocks[l / sqr].st;
      for (auto o : blocks[l / sqr].dq) {
        if (idx == l) st.push_back(val);
        st.push_back(o);
        idx++;
      }
      blocks[l / sqr].cnt[val]++;
      blocks[r / sqr].cnt[val]--;
      blocks[l / sqr].dq.clear();
      blocks[r / sqr].dq.clear();
      for (auto o : st) blocks[l / sqr].dq.push_back(o);
      for (auto o : en) blocks[r / sqr].dq.push_back(o);
      while (l / sqr < r / sqr) {
        int tmp = blocks[l / sqr].dq.back();
        blocks[l / sqr].dq.pop_back();
        blocks[l / sqr].cnt[tmp]--;
        l += sqr;
        blocks[l / sqr].dq.push_front(tmp);
        blocks[l / sqr].cnt[tmp]++;
      }
    }
  }
}