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 N = 1e5 + 50;
const long long oo = 1e18;
const int mod = 1e9 + 7;
int Z, sum[400][N];
deque<int> q[400];
void add(int l, int r) {
  if (l == r) return;
  int st = l / Z;
  int en = r / Z;
  if (st == en) {
    vector<int> tmp;
    while (q[st].size()) {
      tmp.push_back(q[st].front());
      sum[st][q[st].front()]--;
      q[st].pop_front();
    }
    for (int i = 0; i < tmp.size(); i++) {
      if (i == r - st * Z) continue;
      if (i == l - st * Z) {
        sum[st][tmp[r - st * Z]]++;
        q[st].push_back(tmp[r - st * Z]);
      }
      sum[st][tmp[i]]++;
      q[st].push_back(tmp[i]);
    }
  } else {
    int fr = l / Z + 1, to = r / Z - 1;
    int idx = fr;
    int add = q[to].back();
    while (idx <= to) {
      sum[idx][q[idx - 1].back()]++;
      q[idx].push_front(q[idx - 1].back());
      idx++;
    }
    idx = fr;
    while (idx <= to) {
      sum[idx][q[idx].back()]--;
      q[idx].pop_back();
      idx++;
    }
    vector<int> ml;
    while (q[st].size()) {
      sum[st][q[st].front()]--;
      ml.push_back(q[st].front());
      q[st].pop_front();
    }
    vector<int> mr;
    while (q[en].size()) {
      sum[en][q[en].front()]--;
      mr.push_back(q[en].front());
      q[en].pop_front();
    }
    for (int i = 0; i < ml.size(); ++i) {
      if (i == l - st * Z) {
        sum[st][mr[r - en * Z]]++;
        q[st].push_back(mr[r - en * Z]);
      }
      if (i == ml.size() - 1) continue;
      sum[st][ml[i]]++;
      q[st].push_back(ml[i]);
    }
    for (int i = 0; i < mr.size(); ++i) {
      if (!i) {
        q[en].push_back(add);
        sum[en][add]++;
      }
      if (i == r - en * Z) continue;
      sum[en][mr[i]]++;
      q[en].push_back(mr[i]);
    }
  }
}
int main() {
  int n, p = 0;
  cin >> n;
  Z = sqrt(n) + 1;
  for (int i = 0; i < n; i++) {
    int a;
    scanf("%d", &a);
    q[i / Z].push_back(a);
    sum[i / Z][a]++;
  }
  int w;
  int lst = 0;
  cin >> w;
  while (w--) {
    int t, l, r, k;
    scanf("%d%d%d", &t, &l, &r);
    r--;
    l--;
    l += lst;
    r += lst;
    l %= n;
    r %= n;
    if (l > r) swap(l, r);
    if (t == 1) {
      add(l, r);
    } else {
      scanf("%d", &k);
      k = (k + lst - 1) % n + 1;
      lst = 0;
      for (int i = l / Z + 1; i < r / Z; i++) lst += sum[i][k];
      vector<int> tmp;
      while (q[l / Z].size()) {
        int idx = tmp.size() + (l / Z) * Z;
        if (idx >= l && idx <= r && q[l / Z].front() == k) lst++;
        tmp.push_back(q[l / Z].front());
        q[l / Z].pop_front();
      }
      for (int i = 0; i < tmp.size(); ++i) {
        q[l / Z].push_back(tmp[i]);
      }
      if (l / Z != r / Z) {
        vector<int> tmp2;
        while (q[r / Z].size()) {
          int idx = tmp2.size() + (r / Z) * Z;
          if (idx >= l && idx <= r && q[r / Z].front() == k) lst++;
          tmp2.push_back(q[r / Z].front());
          q[r / Z].pop_front();
        }
        for (int i = 0; i < tmp2.size(); ++i) {
          q[r / Z].push_back(tmp2[i]);
        }
      }
      printf("%d\n", lst);
    }
  }
}