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 long long mod(long long n, long long m) {
  long long ret = n % m;
  if (ret < 0) ret += m;
  return ret;
}
long long gcd(long long a, long long b) {
  return (b == 0LL ? a : gcd(b, a % b));
}
long long exp(long long a, long long b, long long m) {
  if (b == 0LL) return 1LL;
  if (b == 1LL) return mod(a, m);
  long long k = mod(exp(a, b / 2, m), m);
  if (b & 1LL) {
    return mod(a * mod(k * k, m), m);
  } else
    return mod(k * k, m);
}
int n;
const int N = 100100;
const int sq = 350;
int cnt[N][N / sq];
vector<int> block[N / sq];
int lastBlock = 0;
void getId(int& b, int& id2, int id) {
  b = id2 = 0;
  while (b < lastBlock and id >= block[b].size()) {
    id -= block[b++].size();
  }
  id2 = id;
}
void insert(int id, int val) {
  int b, id2;
  getId(b, id2, id);
  cnt[val][b]++;
  block[b].insert(block[b].begin() + id2, val);
}
void erase(int id, int val) {
  int b, id2;
  getId(b, id2, id);
  cnt[val][b]--;
  assert((int)block[b].size() > id2);
  block[b].erase(block[b].begin() + id2);
}
void rebuild() {
  vector<int> v;
  for (int i = 0; i <= lastBlock; i++) {
    for (int x : block[i]) {
      cnt[x][i] = 0;
      v.push_back(x);
    }
    block[i].clear();
  }
  for (int i = 0; i < (int)v.size(); i++) {
    int b = i / sq;
    block[b].push_back(v[i]);
    cnt[v[i]][b]++;
  }
}
int query(int x, int k) {
  if (x < 0) return 0;
  int res = 0;
  int b, id;
  getId(b, id, x);
  for (int i = 0; i < b; i++) {
    res += cnt[k][i];
  }
  for (int i = 0; i <= id; i++) {
    res += (block[b][i] == k);
  }
  return res;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    block[i / sq].push_back(x);
  }
  lastBlock = (n - 1) / sq;
  int q;
  cin >> q;
  int last = 0;
  for (int it = 0; it < q; it++) {
    if (it % sq == 0) {
      rebuild();
    }
    int op;
    cin >> op;
    int l, r;
    cin >> l >> r;
    l = mod(l + last - 1, n);
    r = mod(r + last - 1, n);
    if (l > r) swap(l, r);
    if (op == 1) {
      int b, id;
      getId(b, id, r);
      int val = block[b][id];
      erase(r, val);
      insert(l, val);
    } else {
      int k;
      cin >> k;
      k = mod(k + last - 1, n) + 1;
      last = query(r, k) - query(l - 1, k);
      cout << last << "\n";
    }
  }
}