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 = 100010;
deque<int> cur[320];
int freq[320][N];
int n, q, Z, lastAns = 0, a[N], type, l, r, k;
void fix(int l, int r) {
  if (l == r) return;
  int myL = l / Z;
  int myR = r / Z;
  if (myL == myR) {
    vector<int> tmp;
    while (cur[myL].size()) {
      tmp.push_back(cur[myL].front());
      freq[myL][cur[myL].front()]--;
      cur[myL].pop_front();
    }
    for (int i = 0; i < tmp.size(); ++i) {
      if (i == r - myL * Z) continue;
      if (i == l - myL * Z) {
        freq[myL][tmp[r - myL * Z]]++;
        cur[myL].push_back(tmp[r - myL * Z]);
      }
      freq[myL][tmp[i]]++;
      cur[myL].push_back(tmp[i]);
    }
  } else {
    int from = l / Z + 1, to = r / Z - 1;
    int idx = from;
    int toAdd = cur[to].back();
    while (idx <= to) {
      freq[idx][cur[idx - 1].back()]++;
      cur[idx].push_front(cur[idx - 1].back());
      idx++;
    }
    idx = from;
    while (idx <= to) {
      freq[idx][cur[idx].back()]--;
      cur[idx].pop_back();
      idx++;
    }
    vector<int> ml;
    while (cur[myL].size()) {
      freq[myL][cur[myL].front()]--;
      ml.push_back(cur[myL].front());
      cur[myL].pop_front();
    }
    vector<int> mr;
    while (cur[myR].size()) {
      freq[myR][cur[myR].front()]--;
      mr.push_back(cur[myR].front());
      cur[myR].pop_front();
    }
    for (int i = 0; i < ml.size(); ++i) {
      if (i == l - myL * Z) {
        freq[myL][mr[r - myR * Z]]++;
        cur[myL].push_back(mr[r - myR * Z]);
      }
      if (i == ml.size() - 1) continue;
      freq[myL][ml[i]]++;
      cur[myL].push_back(ml[i]);
    }
    for (int i = 0; i < mr.size(); ++i) {
      if (!i) {
        cur[myR].push_back(toAdd);
        freq[myR][toAdd]++;
      }
      if (i == r - myR * Z) continue;
      freq[myR][mr[i]]++;
      cur[myR].push_back(mr[i]);
    }
  }
}
int count(deque<int> &q, int k, int l, int r, int block) {
  int res = 0;
  vector<int> tmp;
  while (q.size()) {
    int curIdx = tmp.size() + block * Z;
    if (curIdx >= l && curIdx <= r && q.front() == k) res++;
    tmp.push_back(q.front());
    q.pop_front();
  }
  for (int i = 0; i < tmp.size(); ++i) {
    q.push_back(tmp[i]);
  }
  return res;
}
int main() {
  scanf("%d", &n);
  Z = sqrt(n) + 1;
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
    cur[i / Z].push_back(a[i]);
    freq[i / Z][a[i]]++;
  }
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d%d", &type, &l, &r);
    l--, r--;
    l = l + lastAns;
    r = r + lastAns;
    l %= n;
    r %= n;
    if (l > r) swap(l, r);
    if (type == 1) {
      fix(l, r);
    } else {
      scanf("%d", &k);
      k = (k + lastAns - 1) % n + 1;
      int res = 0;
      int from = l / Z + 1, to = r / Z - 1;
      while (from <= to) {
        res += freq[from][k];
        from++;
      }
      int myL = l / Z;
      int myR = r / Z;
      res += count(cur[myL], k, l, r, myL);
      if (myL != myR) res += count(cur[myR], k, l, r, myR);
      printf("%d\n", res);
      lastAns = res;
    }
  }
  return 0;
}