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
#include <bits/stdc++.h>
using namespace std;
namespace zzc {
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x * f;
}
const int maxn = 1e5 + 5;
const int sqr = 320;
int n, blo, cnt = 1, qt, lst = 0;
int num[sqr][maxn], bel[maxn], lef[maxn], rig[maxn], a[maxn];
list<int> li[sqr];
void work() {
  int opt, l, r, x;
  n = read();
  blo = 600;
  for (int i = 1; i <= n; i++) a[i] = read();
  for (int i = 1, l = 1, r; l <= n; l = r + 1, i++) {
    r = min(n, l + blo - 1);
    lef[i] = l;
    rig[i] = r;
    for (int j = l; j <= r; j++) {
      bel[j] = i;
      ++num[i][a[j]];
      li[i].push_back(a[j]);
    }
  }
  qt = read();
  while (qt--) {
    opt = read();
    l = (read() + lst - 1) % n + 1;
    r = (read() + lst - 1) % n + 1;
    if (l > r) swap(l, r);
    if (opt == 1) {
      auto it = li[bel[r]].begin();
      for (int i = lef[bel[r]]; i < r; i++) it++;
      int tmp = (*it);
      num[bel[r]][tmp]--;
      num[bel[l]][tmp]++;
      li[bel[r]].erase(it);
      it = li[bel[l]].begin();
      for (int i = lef[bel[l]]; i < l; i++) it++;
      li[bel[l]].insert(it, tmp);
      for (int i = bel[l]; i < bel[r]; i++) {
        tmp = li[i].back();
        li[i].pop_back();
        li[i + 1].push_front(tmp);
        num[i][tmp]--;
        num[i + 1][tmp]++;
      }
    } else if (opt == 2) {
      x = (read() + lst - 1) % n + 1;
      if (bel[l] == bel[r]) {
        lst = 0;
        auto it = li[bel[l]].begin();
        for (int i = lef[bel[l]]; i <= r; i++, it++)
          if (i >= l) lst += (*it) == x;
      } else {
        lst = 0;
        auto it = li[bel[l]].begin();
        for (int i = lef[bel[l]]; i <= rig[bel[l]]; i++, it++)
          if (i >= l) lst += (*it) == x;
        it = li[bel[r]].begin();
        for (int i = lef[bel[r]]; i <= r; i++, it++) lst += (*it) == x;
        for (int i = bel[l] + 1; i < bel[r]; i++) lst += num[i][x];
      }
      printf("%d\n", lst);
    }
  }
}
}  // namespace zzc
int main() {
  zzc::work();
  return 0;
}
