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 sum = 0, ff = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') ff = -1;
    ch = getchar();
  }
  while (isdigit(ch)) sum = sum * 10 + (ch ^ 48), ch = getchar();
  return sum * ff;
}
const int mod = 1e9 + 7;
const int mo = 998244353;
const int N = 1e5 + 5;
const int M = 400;
int n, m, siz, link[N], c[M][N], L[N], R[N];
deque<int> a[N];
int main() {
  n = read();
  siz = pow(n, 0.618);
  for (int i = (1); i <= (n); i++) {
    int x = read(), y;
    link[i] = y = (i - 1) / siz + 1;
    a[y].push_back(x);
    c[y][x]++;
  }
  m = (n - 1) / siz + 1;
  for (int i = (1); i <= (m); i++) {
    L[i] = (i - 1) * siz + 1;
    R[i] = i * siz;
  }
  R[m] = n;
  int Q = read();
  int lasans = 0;
  for (int q = (1); q <= (Q); q++) {
    int op, x, y, z;
    op = read();
    if (op == 1) {
      x = read();
      y = read();
      x = (x + lasans - 1) % n + 1;
      y = (y + lasans - 1) % n + 1;
      if (x > y) swap(x, y);
      int p = link[x], q = link[y];
      int f = x - L[p];
      int g = y - L[q];
      int v = a[q][g];
      if (p == q) {
        a[q].erase(a[q].begin() + g);
        a[p].insert(a[p].begin() + f, v);
      } else {
        a[p].insert(a[p].begin() + f, v);
        a[q].erase(a[q].begin() + g);
        c[p][v]++;
        c[q][v]--;
        for (int j = (p); j <= (q - 1); j++) {
          int v = a[j].back();
          a[j].pop_back();
          c[j][v]--;
          a[j + 1].push_front(v);
          c[j + 1][v]++;
        }
      }
    } else {
      x = read();
      y = read();
      z = read();
      x = (x + lasans - 1) % n + 1;
      y = (y + lasans - 1) % n + 1;
      if (x > y) swap(x, y);
      z = (z + lasans - 1) % n + 1;
      int p = link[x], q = link[y];
      int f = x - L[p];
      int g = y - L[q];
      int ans = 0;
      if (p == q) {
        for (int j = (f); j <= (g); j++) ans += a[p][j] == z;
      } else {
        for (int j = (f); j <= (a[p].size() - 1); j++) ans += a[p][j] == z;
        for (int j = (0); j <= (g); j++) ans += a[q][j] == z;
        for (int j = (p + 1); j <= (q - 1); j++) ans += c[j][z];
      }
      printf("%d\n", ans);
      lasans = ans;
    }
  }
  return 0;
}