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 maxn = 1e5 + 233;
const int siz = 500;
const int maxs = 2000;
const int tot = maxn / siz + 5;
int tmp[maxn], num, c[tot][maxn], s[tot];
vector<int> a[tot];
int ans = 0, n;
void rebuild() {
  int j;
  for (int i = 0; i < num; i++) {
    s[i] = 0;
    a[i].clear();
    a[i].reserve(siz);
  }
  for (int i = 0; i < n; i++) {
    int j = i / siz;
    a[j].push_back(tmp[i]);
    c[j][tmp[i]]++;
    s[j]++;
  }
  return;
}
void pre() {
  int cur = 0;
  for (int i = 0; i < num; i++)
    for (int j = 0; j < s[i]; j++) tmp[cur++] = a[i][j], c[i][a[i][j]]--;
  return;
}
void dcode(int &x) {
  x = (ans + x - 1) % n + 1;
  return;
}
int main() {
  cin >> n;
  num = (n + siz - 1) / siz;
  for (int i = 0; i < n; i++) scanf("%d", &tmp[i]);
  rebuild();
  int q;
  cin >> q;
  while (q--) {
    int badnode = 0;
    int op, l, r, k;
    scanf("%d%d%d", &op, &l, &r);
    dcode(l);
    dcode(r);
    l--;
    r--;
    if (l > r) swap(l, r);
    if (op == 1) {
      int L = 0, R = 0, buf;
      for (int i = 0; i < num; i++) {
        R = L + s[i] - 1;
        if (L <= r && r <= R) {
          buf = a[i][r - L];
          c[i][buf]--;
          s[i]--;
          a[i].erase(a[i].begin() + r - L);
          break;
        }
        L += s[i];
      }
      L = 0;
      R = 0;
      for (int i = 0; i < num; i++) {
        R = L + s[i] - 1;
        if (L <= l && l <= R) {
          c[i][buf]++;
          s[i]++;
          a[i].insert(a[i].begin() + l - L, buf);
          if (s[i] == maxs) badnode = 1;
          break;
        }
        L += s[i];
      }
    } else {
      scanf("%d", &k);
      dcode(k);
      ans = 0;
      int L = 0, R = 0;
      for (int i = 0; i < num; i++) {
        R = L + s[i] - 1;
        if (R < l)
          ;
        else if (L > r)
          break;
        else if (l <= L && R <= r)
          ans += c[i][k];
        else {
          int lb = max(l, L) - L, rb = min(r, R) - L;
          for (int j = lb; j <= rb; j++)
            if (a[i][j] == k) ans++;
        }
        L += s[i];
      }
      printf("%d\n", ans);
    }
    if (badnode) {
      pre();
      rebuild();
    }
  }
  return 0;
}