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;
int bloco_tam, n, v[100010];
int cont[320][100010];
deque<int> bloco[320];
void constroi() {
  int id = 0;
  bloco_tam = sqrt(n);
  for (int i = 0; i < n; i++) {
    if (i > 0 && i % bloco_tam == 0) id++;
    cont[id][v[i]]++;
    bloco[id].push_back(v[i]);
  }
}
int get(int l, int r, int val) {
  int res = 0;
  while (l > 0 && l % bloco_tam != 0 && l < r) {
    if (bloco[l / bloco_tam][l % bloco_tam] == val) res++;
    l++;
  }
  while (l + bloco_tam <= r) {
    res += cont[l / bloco_tam][val];
    l += bloco_tam;
  }
  while (l <= r) {
    if (bloco[l / bloco_tam][l % bloco_tam] == val) res++;
    l++;
  }
  return res;
}
void update(int l, int r) {
  int dir = r / bloco_tam;
  int esq = l / bloco_tam;
  int fim = bloco[dir][r % bloco_tam];
  bloco[dir].erase(bloco[dir].begin() + r % bloco_tam);
  cont[dir][fim]--;
  bloco[esq].insert(bloco[esq].begin() + l % bloco_tam, fim);
  cont[esq][fim]++;
  if (esq == dir) return;
  int atu = bloco[esq].back();
  bloco[esq].pop_back();
  cont[esq][atu]--;
  esq++;
  while (esq < dir) {
    bloco[esq].push_front(atu);
    cont[esq][atu]++;
    atu = bloco[esq].back();
    bloco[esq].pop_back();
    cont[esq][atu]--;
    esq++;
  }
  bloco[esq].push_front(atu);
  cont[esq][atu]++;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &v[i]);
  constroi();
  int q;
  scanf("%d", &q);
  int lastans = 0;
  while (q--) {
    int op;
    scanf("%d", &op);
    if (op == 2) {
      int l, r, val;
      scanf("%d %d %d", &l, &r, &val);
      l = (l + lastans - 1) % n + 1;
      r = (r + lastans - 1) % n + 1;
      val = (val + lastans - 1) % n + 1;
      if (l > r) swap(l, r);
      lastans = get(l - 1, r - 1, val);
      printf("%d\n", lastans);
    } else {
      int l, r;
      scanf("%d %d", &l, &r);
      l = (l + lastans - 1) % n + 1;
      r = (r + lastans - 1) % n + 1;
      if (l > r) swap(l, r);
      update(l - 1, r - 1);
    }
  }
  return 0;
}