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 = 100335;
const int len = 633;
int cnt[len][N];
vector<int> b[len];
int a[N];
int main() {
  int n, q, type, l, r, k, lastans = 0, j1, j2;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int r;
    scanf("%d", &r);
    b[i / len].push_back(r);
    cnt[i / len][r]++;
  }
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    if (i % len == 0) {
      int v = 0;
      for (int j = 0; j < len; j++) {
        for (int p = 0; p < b[j].size(); p++) {
          cnt[j][b[j][p]]--;
          a[v++] = b[j][p];
        }
        b[j].clear();
      }
      for (int j = 0; j < n; j++) {
        b[j / len].push_back(a[j]);
        cnt[j / len][a[j]]++;
      }
    }
    scanf("%d", &type);
    if (type == 1) {
      scanf("%d%d", &l, &r);
      l = (l + lastans - 1) % n + 1;
      r = (r + lastans - 1) % n + 1;
      if (l > r) swap(l, r);
      for (j1 = 0, j2 = 0; b[j1].size() < l || b[j2].size() < r;) {
        if (b[j1].size() < l) l -= b[j1++].size();
        if (b[j2].size() < r) r -= b[j2++].size();
      }
      int p = *(b[j2].begin() + r - 1);
      b[j2].erase(b[j2].begin() + r - 1);
      cnt[j2][p]--;
      b[j1].insert(b[j1].begin() + l - 1, p);
      cnt[j1][p]++;
    } else {
      scanf("%d%d%d", &l, &r, &k);
      l = (l + lastans - 1) % n + 1;
      r = (r + lastans - 1) % n + 1;
      k = (k + lastans - 1) % n + 1;
      if (l > r) swap(l, r);
      lastans = 0;
      for (j1 = 0, j2 = 0; b[j1].size() < l || b[j2].size() < r;) {
        if (b[j1].size() < l) l -= b[j1++].size();
        if (b[j2].size() < r) r -= b[j2++].size();
      }
      if (j1 == j2) {
        for (int j = l - 1; j <= r - 1; j++) lastans += (b[j1][j] == k);
      } else {
        for (int j = j1 + 1; j <= j2 - 1; j++) lastans += cnt[j][k];
        for (int j = l - 1; j < b[j1].size(); j++) lastans += (b[j1][j] == k);
        for (int j = 0; j <= r - 1; j++) lastans += (b[j2][j] == k);
      }
      printf("%d\n", lastans);
    }
  }
  return 0;
}