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;
inline int read() {
  char c = getchar();
  int x = 0, f = 1;
  while ((c < '0' || c > '9') && c - '-') c = getchar();
  if (c == '-') f = -1, c = getchar();
  while (c >= '0' && c <= '9') {
    x = (x << 3) + (x << 1) + c - 48;
    c = getchar();
  }
  return x * f;
}
const int N = 1e5 + 10;
int n, a[N], c[355][N], lstans = 0, bel[N], siz;
deque<int> que[355];
int m;
inline void shuchu() {
  for (int i = 1; i <= bel[n]; i++)
    for (int j = 0; j < que[i].size(); j++) cout << que[i][j] << " ";
  cout << "\n";
}
int main() {
  n = read();
  siz = sqrt(n);
  for (int i = 1; i <= n; i++) {
    a[i] = read();
    bel[i] = (i - 1) / siz + 1;
    que[bel[i]].push_back(a[i]);
    c[bel[i]][a[i]]++;
  }
  m = read();
  for (int o = 1, typ, l, r, k; o <= m; o++) {
    typ = read();
    l = (read() + lstans - 1) % n + 1;
    r = (read() + lstans - 1) % n + 1;
    if (l > r) swap(l, r);
    if (typ == 1) {
      int xl = l - (bel[l] - 1) * siz, xr = r - (bel[r] - 1) * siz;
      if (bel[l] == bel[r]) {
        int be = bel[r];
        int zhi = que[be][xr - 1];
        que[be].erase(que[be].begin() + xr - 1);
        if (!que[be].empty())
          que[be].insert(que[be].begin() + xl - 1, zhi);
        else
          que[be].push_back(zhi);
        continue;
      }
      int x = que[bel[r]][xr - 1], ji = que[bel[l]].back();
      c[bel[r]][x]--;
      que[bel[r]].erase(que[bel[r]].begin() + xr - 1);
      c[bel[l]][ji]--;
      que[bel[l]].pop_back();
      c[bel[l]][x]++;
      que[bel[l]].insert(que[bel[l]].begin() + xl - 1, x);
      for (int i = bel[l] + 1; i < bel[r]; i++) {
        int pao = que[i].back();
        c[i][pao]--;
        que[i].pop_back();
        c[i][ji]++;
        que[i].push_front(ji);
        ji = pao;
      }
      c[bel[r]][ji]++;
      que[bel[r]].push_front(ji);
    }
    if (typ == 2) {
      k = (read() + lstans - 1) % n + 1;
      int dan = 0;
      if (bel[l] == bel[r]) {
        int be = bel[l], xl = l - (bel[l] - 1) * siz,
            xr = r - (bel[r] - 1) * siz;
        for (int i = xl; i <= xr; i++)
          if (que[be][i - 1] == k) dan++;
      } else {
        int xl = l - (bel[l] - 1) * siz;
        while ((l % siz) ^ 1) {
          if (que[bel[l]][xl - 1] == k) dan++;
          l++;
          xl++;
        }
        int xr = r - (bel[r] - 1) * siz;
        while (r % siz) {
          if (que[bel[r]][xr - 1] == k) dan++;
          r--;
          xr--;
        }
        int zuo = bel[l], you = bel[r];
        for (int i = zuo; i <= you; i++) dan += c[i][k];
      }
      cout << (lstans = dan) << "\n";
    }
  }
  return 0;
}