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 mod = 1000000007, N = 500005;
long long a, b, c, d, e, g, arr[N], lastans;
deque<long long> q[1005];
long long mpi[320][100005];
vector<pair<long long, long long> > v;
void reorder() {
  long long x = 0;
  while (v.size()) {
    mpi[v[v.size() - 1].first][v[v.size() - 1].second] = 0;
    v.pop_back();
  }
  for (long long i = 0; i <= ceil((double)a / (long long)sqrt(a)); i++)
    while (q[i].size()) {
      arr[++x] = q[i].front();
      q[i].pop_front();
    }
  x = sqrt(a);
  for (long long i = 1; i <= a; i++) {
    q[(const int)ceil((double)i / x)].push_back(arr[i]);
    if (mpi[(const int)ceil((double)i / x)][arr[i]] == 0)
      v.push_back(make_pair(ceil((double)i / x), arr[i]));
    mpi[(const int)ceil((double)i / x)][arr[i]]++;
  }
}
void sorgu_1() {
  long long z, y, sum = 0, zz, zzz, founded;
  cin >> z >> y;
  z = ((z + lastans - 1) % a) + 1;
  y = ((y + lastans - 1) % a) + 1;
  if (z > y) swap(z, y);
  for (long long i = 1;; i++) {
    if (z > sum and sum + q[i].size() >= z) zz = i, zzz = sum;
    if (sum + q[i].size() >= y) {
      long long s = y - sum;
      stack<long long> stc;
      while (--s) {
        stc.push(q[i].front());
        q[i].pop_front();
      }
      founded = q[i].front();
      mpi[i][founded]--;
      mpi[zz][founded]++;
      if (mpi[zz][founded] == 1) v.push_back(make_pair(zz, founded));
      q[i].pop_front();
      while (stc.size()) {
        q[i].push_front(stc.top());
        stc.pop();
      }
      break;
    }
    sum += q[i].size();
  }
  long long s = z - zzz;
  stack<long long> stc;
  while (--s) {
    stc.push(q[zz].front());
    q[zz].pop_front();
  }
  q[zz].push_front(founded);
  while (stc.size()) {
    q[zz].push_front(stc.top());
    stc.pop();
  }
}
void sorgu_2() {
  long long y = 0, z = 0, t = 0, ans = 0, sum = 0;
  cin >> z >> y >> t;
  z = ((z + lastans - 1) % a) + 1;
  y = ((y + lastans - 1) % a) + 1;
  t = ((t + lastans - 1) % a) + 1;
  if (z > y) swap(z, y);
  bool flag = 0;
  for (long long i = 1;; i++) {
    if (sum + q[i].size() >= y and sum + q[i].size() >= z and z > sum) {
      long long s = z - sum;
      stack<long long> stc;
      while (--s) {
        stc.push(q[i].front());
        q[i].pop_front();
      }
      s = y - z + 1;
      while (s--) {
        if (q[i].front() == t) ans++;
        stc.push(q[i].front());
        q[i].pop_front();
      }
      while (stc.size()) {
        q[i].push_front(stc.top());
        stc.pop();
      }
      break;
    }
    if (sum + q[i].size() >= y) {
      long long s = y - sum;
      stack<long long> stc;
      while (s--) {
        if (q[i].front() == t) ans++;
        stc.push(q[i].front());
        q[i].pop_front();
      }
      while (stc.size()) {
        q[i].push_front(stc.top());
        stc.pop();
      }
      break;
    }
    if (flag == 1) {
      ans += mpi[i][t];
    }
    if (sum + q[i].size() >= z and z > sum) {
      long long s = sum + q[i].size() - z + 1;
      stack<long long> stc;
      while (s--) {
        if (q[i].back() == t) ans++;
        stc.push(q[i].back());
        q[i].pop_back();
      }
      while (stc.size()) {
        q[i].push_back(stc.top());
        stc.pop();
      }
      flag = 1;
    }
    sum += q[i].size();
  }
  lastans = ans;
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> a;
  long long x = sqrt(a);
  for (long long i = 1; i <= a; i++) {
    cin >> arr[i];
    q[(const int)ceil((double)i / x)].push_back(arr[i]);
    if (mpi[(const int)ceil((double)i / x)][arr[i]] == 0)
      v.push_back(make_pair(ceil((double)i / x), arr[i]));
    mpi[(const int)ceil((double)i / x)][arr[i]]++;
  }
  cin >> g;
  long long say = 0;
  x += 500;
  while (g--) {
    long long t = 0;
    cin >> t;
    if (t == 1) {
      say++;
      sorgu_1();
      if (say % x == 0) reorder();
    } else {
      sorgu_2();
    }
  }
}