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>
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const int N = 1e5 + 10;
const int sz = 320;
int a[N], n, Q;
deque<int> q[320];
int cnt[320][N];
void update(int l, int r) {
  l--, r--;
  int lb = l / sz;
  int rb = r / sz;
  int v = q[rb][r % sz];
  q[rb].erase(q[rb].begin() + (r % sz));
  q[lb].insert(q[lb].begin() + (l % sz), v);
  cnt[rb][v]--;
  cnt[lb][v]++;
  while (lb < rb) {
    v = q[lb].back();
    q[lb].pop_back();
    cnt[lb][v]--;
    q[++lb].push_front(v);
    cnt[lb][v]++;
  }
}
int query(int l, int r, int k) {
  l--, r--;
  int lb = l / sz;
  int rb = r / sz;
  if (lb == rb) {
    int ret = 0;
    l %= sz;
    r %= sz;
    for (int i = l; i <= r; i++) ret += q[lb][i] == k;
    return ret;
  }
  int ret = 0;
  for (int i = l % sz; i < q[lb].size(); i++) ret += q[lb][i] == k;
  int lst = r % sz;
  for (int i = 0; i <= lst; i++) ret += q[rb][i] == k;
  for (int i = lb + 1; i < rb; i++) ret += cnt[i][k];
  return ret;
}
void print() {
  for (int i = 0;; i++) {
    if (q[i].size()) {
      for (int x : q[i]) cout << x << " ";
    } else
      break;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    cnt[i / sz][a[i]]++;
    q[i / sz].push_back(a[i]);
  }
  scanf("%d", &Q);
  int last = 0;
  while (Q--) {
    int t, l, r, k;
    scanf("%d %d %d", &t, &l, &r);
    l = (l + last - 1) % n + 1;
    r = (r + last - 1) % n + 1;
    if (l > r) swap(l, r);
    if (t == 1) {
      update(l, r);
    } else {
      scanf("%d", &k);
      k = (k + last - 1) % n + 1;
      printf("%d\n", last = query(l, r, k));
    }
  }
}