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() {
  int num = 0;
  char c = getchar();
  if (c == '-') return -read();
  while (c >= '0' && c <= '9') {
    num = (num << 3) + (num << 1) + c - '0';
    c = getchar();
  }
  return num;
}
inline long long READ() {
  long long num = 0;
  char c = getchar();
  if (c == '-') return -READ();
  while (c >= '0' && c <= '9') {
    num = (num << 3) + (num << 1) + c - '0';
    c = getchar();
  }
  return num;
}
deque<int> dq[357];
int n, m, q, mode, L, R, l, r, ans = 0, bb, bs;
int hvn[357][100007];
void deal(int& a) { a = (a + ans - 1) % n + 1; }
int main() {
  scanf("%d", &n);
  bs = int(sqrt(n)) + 1;
  for (int i = 0; i < n; i++) {
    scanf("%d", &bb);
    dq[i / bs].push_back(bb);
    hvn[i / bs][bb]++;
  }
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    scanf("%d", &mode);
    if (mode == 1) {
      scanf("%d%d", &L, &R);
      deal(L);
      deal(R);
      L--;
      R--;
      if (L > R) swap(L, R);
      l = L / bs;
      L %= bs;
      r = R / bs;
      R %= bs;
      if (l == r) {
        int x = dq[l][R];
        dq[l].erase(dq[l].begin() + R);
        dq[l].insert(dq[l].begin() + L, x);
      } else {
        int x = dq[r][R];
        dq[r].erase(dq[r].begin() + R);
        hvn[r][x]--;
        for (int i = l; i < r; i++) {
          int y = dq[i].back();
          dq[i].pop_back();
          hvn[i][y]--;
          hvn[i + 1][y]++;
          dq[i + 1].push_front(y);
        }
        dq[l].insert(dq[l].begin() + L, x);
        hvn[l][x]++;
      }
    } else {
      int x;
      scanf("%d%d%d", &L, &R, &x);
      deal(L);
      deal(R);
      deal(x);
      L--;
      R--;
      if (L > R) swap(L, R);
      l = L / bs;
      L %= bs;
      r = R / bs;
      R %= bs;
      ans = 0;
      if (l == r) {
        for (int i = L; i <= R; i++)
          if (dq[l][i] == x) ans++;
      } else {
        for (int i = l + 1; i < r; i++) ans += hvn[i][x];
        for (int i = L; i < bs; i++)
          if (dq[l][i] == x) ans++;
        for (int i = R; i >= 0; i--)
          if (dq[r][i] == x) ans++;
      }
      printf("%d\n", ans);
    }
  }
  return 0;
}