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;
int n, m, ans = 0;
vector<int> g[400];
int buc[350][100005];
int a[100005];
int main() {
  scanf("%d", &n);
  int div = sqrt(n), m = n / div + 1;
  memset(buc, 0, sizeof(buc));
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    a[i]--;
  }
  for (int i = 0; i < m; i++) g[i].clear();
  for (int i = 0; i < n; i++) {
    g[i / div].push_back(a[i]);
    buc[i / div][a[i]]++;
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int typ, l, r;
    scanf("%d%d%d", &typ, &l, &r);
    l = (l + ans - 1) % n;
    r = (r + ans - 1) % n;
    if (l > r) swap(l, r);
    bool f = false;
    if (typ == 1) {
      int x;
      int sum = 0;
      for (int i = 0; i < m; i++) {
        if (sum <= r && r < sum + g[i].size()) {
          x = g[i][r - sum];
          g[i].erase(g[i].begin() + (r - sum));
          buc[i][x]--;
          if (g[i].size() > 4 * div) f = true;
          break;
        } else
          sum += g[i].size();
      }
      sum = 0;
      for (int i = 0; i < m; i++) {
        if (sum <= l && l <= sum + g[i].size()) {
          g[i].insert(g[i].begin() + (l - sum), x);
          buc[i][x]++;
          if (g[i].size() > 4 * div) f = true;
          break;
        } else
          sum += g[i].size();
      }
    } else if (typ == 2) {
      int k;
      scanf("%d", &k);
      k = (k + ans - 1) % n;
      ans = 0;
      int sum = 0;
      for (int i = 0; i < m; i++) {
        int xx = sum + g[i].size() - 1;
        if (xx < l || sum > r)
          sum = xx + 1;
        else if (l <= sum && xx <= r) {
          ans += buc[i][k];
          sum = xx + 1;
        } else {
          int x1 = max(l, sum), x2 = min(r, xx);
          for (int j = x1; j <= x2; j++) ans += (g[i][j - sum] == k);
          if (g[i].size() > 4 * div) f = true;
          sum = xx + 1;
        }
      }
      printf("%d\n", ans);
    }
    if (f) {
      int num = 0;
      for (int i = 0; i < m; i++)
        for (int j = 0; j < g[i].size(); j++) a[num++] = g[i][j];
      for (int i = 0; i < m; i++) {
        for (int j = 0; j < g[i].size(); j++) {
          buc[i][g[i][j]]--;
        }
        g[i].clear();
      }
      for (int i = 0; i < n; i++) {
        g[i / div].push_back(a[i]);
        buc[i / div][a[i]]++;
      }
    }
  }
  return 0;
}