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 B = 320;
const int N_ = 100005;
const int Q_ = 100005;
int N, Q;
int C[B + 12][N_];
int A[N_];
int bk[B + 12][B + B + 12], sz[B + 12];
int pos[N_];
void find_loc(int x, int &b, int &w) {
  int s = 0;
  for (int i = 0; i < B; i++) {
    if (s + sz[i] > x) {
      b = i;
      w = x - s;
      return;
    }
    s += sz[i];
  }
}
int main() {
  scanf("%d", &N);
  for (int i = 0; i < N; i++) {
    scanf("%d", A + i);
    ++C[i / B][A[i]];
    bk[i / B][sz[i / B]++] = i;
  }
  scanf("%d", &Q);
  int lastans = 0;
  while (Q--) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      int l, r;
      scanf("%d%d", &l, &r);
      l = (l + lastans - 1) % N;
      r = (r + lastans - 1) % N;
      if (l > r) swap(l, r);
      int lb, lw, rb, rw;
      find_loc(l, lb, lw);
      int lv = bk[lb][lw];
      find_loc(r, rb, rw);
      int rv = bk[rb][rw];
      for (int i = rw; i + 1 < sz[rb]; i++) bk[rb][i] = bk[rb][i + 1];
      bk[rb][--sz[rb]] = 0;
      --C[rb][A[rv]];
      for (int i = sz[lb]++; i > lw; i--) bk[lb][i] = bk[lb][i - 1];
      bk[lb][lw] = rv;
      ++C[lb][A[rv]];
    } else if (type == 2) {
      int l, r, k;
      scanf("%d%d%d", &l, &r, &k);
      l = (l + lastans - 1) % N;
      r = (r + lastans - 1) % N;
      if (l > r) swap(l, r);
      k = (k + lastans - 1) % N + 1;
      int ret = 0;
      int lb, lw, rb, rw;
      find_loc(l, lb, lw);
      int lv = bk[lb][lw];
      find_loc(r, rb, rw);
      int rv = bk[rb][rw];
      if (lb == rb) {
        for (int i = lw; i <= rw; i++)
          if (A[bk[lb][i]] == k) ++ret;
      } else {
        for (int i = lw; i < sz[lb]; i++)
          if (A[bk[lb][i]] == k) ++ret;
        for (int i = lb + 1; i <= rb - 1; i++) ret += C[i][k];
        for (int i = 0; i <= rw; i++)
          if (A[bk[rb][i]] == k) ++ret;
      }
      printf("%d\n", ret);
      lastans = ret;
    }
    if (Q % B == 0) {
      int an = 0;
      for (int i = 0; i < B; i++) {
        for (int j = 0; j < sz[i]; j++) {
          pos[an++] = bk[i][j];
          C[i][A[bk[i][j]]] = 0;
        }
        sz[i] = 0;
      }
      for (int i = 0; i < N; i++) {
        int b = i / B;
        bk[b][sz[b]++] = pos[i];
        ++C[b][A[pos[i]]];
      }
    }
  }
  return 0;
}