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;
const int S = 700;
const int B = 145;
int a[N];
int cnt[B][N];
vector<int> b[B];
pair<int, int> find(int pos) {
  int cur = 0;
  for (int i = 0; i < B; ++i) {
    if (cur + int(b[i].size()) <= pos)
      cur += b[i].size();
    else
      return {i, pos - cur};
  }
  assert(0);
  return {-1, -1};
}
void shift(int l, int r) {
  if (l == r) return;
  int i, d, v;
  tie(i, d) = find(r);
  v = b[i][d];
  --cnt[i][v];
  b[i].erase(b[i].begin() + d);
  tie(i, d) = find(l);
  ++cnt[i][v];
  b[i].insert(b[i].begin() + d, v);
}
void rebuild(int n) {
  int p = 0;
  for (int i = 0; i < B; ++i) {
    for (int& v : b[i]) {
      --cnt[i][v];
      a[p++] = v;
    }
    b[i].clear();
  }
  for (int i = 0; i < n; ++i) {
    b[i / S].emplace_back(a[i]);
    ++cnt[i / S][a[i]];
  }
}
int get(int l, int r, int k) {
  auto pl = find(l);
  auto pr = find(r);
  int res = 0;
  if (pl.first == pr.first) {
    for (int i = pl.second; i <= pr.second; ++i) res += b[pl.first][i] == k;
  } else {
    for (int i = pl.second; i < int(b[pl.first].size()); ++i)
      res += b[pl.first][i] == k;
    for (int i = pl.first + 1; i < pr.first; ++i) res += cnt[i][k];
    for (int i = 0; i <= pr.second; ++i) res += b[pr.first][i] == k;
  }
  return res;
}
int main() {
  int n, m, i, qt, l, r, k;
  scanf("%d", &n);
  for (i = 0; i < n; ++i) {
    scanf("%d", a + i);
  }
  rebuild(n);
  scanf("%d", &m);
  int res = 0, c = 0;
  while (m--) {
    scanf("%d %d %d", &qt, &l, &r);
    l += res + n - 1, l %= n;
    r += res + n - 1, r %= n;
    if (l > r) swap(l, r);
    if (qt == 1) {
      shift(l, r);
      ++c;
      if (c == S) {
        rebuild(n);
        c = 0;
      }
    } else {
      scanf("%d", &k);
      k += res + n - 1, k %= n, ++k;
      res = get(l, r, k);
      printf("%d\n", res);
    }
  }
  return 0;
}