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 maxn = 1e5 + 100;
const int radikal = 400;
int cnt, n;
int a[maxn];
vector<int> v[maxn];
int dp[radikal][maxn];
int get(int l, int r, int k) {
  int tmp = 0;
  int ret = 0;
  for (int i = 1; i <= cnt; i++) {
    if (tmp >= l and tmp + v[i].size() <= r) {
      ret += dp[i][k];
      tmp += v[i].size();
    } else if (tmp < r or tmp + v[i].size() >= l) {
      for (auto w : v[i]) {
        tmp++;
        if (tmp >= l and tmp <= r) ret += (w == k);
      }
    } else
      tmp += v[i].size();
  }
  return ret;
}
int del(int x, int idx) {
  for (int i = idx; i < int(v[x].size()) - 1; i++) swap(v[x][i], v[x][i + 1]);
  int k = v[x].back();
  v[x].pop_back();
  dp[x][k]--;
  return k;
}
void add(int x, int idx, int val) {
  vector<int> g;
  for (int i = 0; i < idx; i++) g.push_back(v[x][i]);
  g.push_back(val);
  for (int i = idx; i < v[x].size(); i++) g.push_back(v[x][i]);
  v[x] = g;
  dp[x][val]++;
}
void shift(int l, int r) {
  int tmp = 0;
  int m;
  for (int i = 1; i <= cnt; i++) {
    if (tmp + v[i].size() >= r) {
      m = del(i, r - tmp - 1);
      break;
    }
    tmp += v[i].size();
  }
  tmp = 0;
  for (int i = 1; i <= cnt; i++) {
    if (tmp + v[i].size() >= l) {
      add(i, l - tmp - 1, m);
      break;
    }
    tmp += v[i].size();
  }
}
void refresh() {
  int tmp = 1;
  for (int i = 1; i <= cnt; i++) {
    for (auto w : v[i]) {
      dp[i][w]--;
      a[tmp++] = w;
    }
    v[i].clear();
  }
  cnt = 1;
  for (int i = 1; i <= n; i++) {
    if (v[cnt].size() == radikal) cnt++;
    v[cnt].push_back(a[i]);
    dp[cnt][a[i]]++;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  int m;
  cin >> m;
  refresh();
  int lastans = 0;
  for (int i = 1; i <= m; i++) {
    if (i % radikal == 0) refresh();
    int type, l, r, x;
    cin >> type >> l >> r;
    l = ((l + lastans - 1) % n) + 1, r = ((r + lastans - 1) % n) + 1;
    if (l > r) swap(l, r);
    if (type == 1)
      shift(l, r);
    else {
      cin >> x;
      x = ((x + lastans - 1) % n) + 1;
      lastans = get(l, r, x);
      cout << lastans << endl;
    }
  }
}