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 SQ = 350;
const int MAX_N = 100000;
deque<int> a[MAX_N / SQ + 10];
int num[MAX_N / SQ + 10][MAX_N + 10];
int N, Q, lastans = 0;
void print() {
  for (int i = 0; i <= N / SQ; i++) {
    for (int j = 0; j < a[i].size(); j++) {
      cout << a[i][j] << ' ';
    }
  }
  cout << endl;
  return;
}
int main() {
  scanf("%d", &N);
  for (int i = 0; i < N; i++) {
    int x = i / SQ, k;
    scanf("%d", &k);
    a[x].push_back(k);
    num[x][k]++;
  }
  scanf("%d", &Q);
  for (int i = 0; i < Q; i++) {
    int type, l, r, k;
    scanf("%d", &type);
    if (type == 1) {
      scanf("%d %d", &l, &r);
      l = (l + lastans - 1) % N;
      r = (r + lastans - 1) % N;
      if (l > r) swap(l, r);
      int lx = l / SQ, ly = l % SQ, rx = r / SQ, ry = r % SQ;
      if (lx == rx) {
        int f = a[rx][ry];
        a[rx].erase(a[rx].begin() + ry);
        a[rx].insert(a[rx].begin() + ly, f);
      } else {
        int f = a[rx][ry];
        a[rx].erase(a[rx].begin() + ry);
        num[rx][f]--;
        for (int j = lx; j < rx; j++) {
          int b = a[j].back();
          a[j].pop_back();
          a[j + 1].push_front(b);
          num[j][b]--;
          num[j + 1][b]++;
        }
        a[lx].insert(a[lx].begin() + ly, f);
        num[lx][f]++;
      }
    } else {
      scanf("%d %d %d", &l, &r, &k);
      l = (l + lastans - 1) % N;
      r = (r + lastans - 1) % N;
      k = (k + lastans - 1) % N + 1;
      if (l > r) swap(l, r);
      int lx = l / SQ, ly = l % SQ, rx = r / SQ, ry = r % SQ;
      int ans = 0;
      if (lx == rx) {
        for (int j = ly; j <= ry; j++)
          if (a[lx][j] == k) ans++;
      } else {
        for (int j = ly; j < SQ; j++)
          if (a[lx][j] == k) ans++;
        for (int j = lx + 1; j < rx; j++) ans += num[j][k];
        for (int j = 0; j <= ry; j++)
          if (a[rx][j] == k) ans++;
      }
      lastans = ans;
      printf("%d\n", ans);
    }
  }
  return 0;
}