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
#include <bits/stdc++.h>
using namespace std;
const int szS = 300;
struct block {
  vector<int> a;
  int cnt[100005];
  int del(int x) {
    int ret = a[x];
    cnt[ret]--;
    for (int i = x; i < (int)a.size() - 1; ++i) swap(a[i], a[i + 1]);
    a.pop_back();
    return ret;
  }
  void insert(int x, int val) {
    cnt[val]++;
    a.push_back(val);
    int pos = (int)a.size() - 1;
    while (pos != x) {
      swap(a[pos], a[pos - 1]);
      pos--;
    }
  }
  void clear(vector<int>& e) {
    for (int i = 0; i < (int)a.size(); ++i) {
      e.push_back(a[i]);
      cnt[a[i]]--;
    }
    a.clear();
  }
  int count(int l, int r, int val) {
    if (r - l + 1 == (int)a.size()) return cnt[val];
    int ret = 0;
    for (int i = l; i < r + 1; ++i) ret += (val == a[i]);
    return ret;
  }
  void write() {
    for (int i = 0; i < (int)a.size(); ++i) {
      cout << a[i] << " ";
    }
    cout << endl;
  }
};
int n, x;
vector<int> e;
block sq[100005 / szS + 5];
void build(vector<int>& e, int val = 1) {
  if (val) {
    e.clear();
    int all = (n - 1) / szS + 1;
    for (int i = 0; i < all; ++i) {
      sq[i].clear(e);
    }
  }
  for (int i = 0; i < n; ++i) {
    int id = i / szS;
    sq[id].insert(i % szS, e[i]);
  }
}
int del(int pos) {
  int all = (n - 1) / szS + 1;
  for (int i = 0; i < all; ++i) {
    if (pos >= sq[i].a.size())
      pos -= sq[i].a.size();
    else {
      return sq[i].del(pos);
    }
  }
  return 0;
}
void add(int pos, int val) {
  int all = (n - 1) / szS + 1;
  for (int i = 0; i < all; ++i) {
    if (pos > sq[i].a.size())
      pos -= sq[i].a.size();
    else {
      sq[i].insert(pos, val);
      return;
    }
  }
  sq[all - 1].insert((int)sq[all - 1].a.size(), val);
}
int get(int l, int r, int val) {
  int ret = 0, all = (n - 1) / szS + 1;
  int ln, rn;
  for (int i = 0; i < all; ++i) {
    if (!i) {
      ln = 0;
      rn = (int)sq[i].a.size() - 1;
    } else {
      ln = rn + 1;
      rn = ln + (int)sq[i].a.size() - 1;
    }
    int lc, rc;
    lc = max(ln, l);
    rc = min(rn, r);
    if (lc > rc) continue;
    ret += sq[i].count(lc - ln, rc - ln, val);
  }
  return ret;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &x);
    e.push_back(x);
  }
  build(e, 0);
  int m;
  scanf("%d", &m);
  int lastans = 0;
  for (int i = 0; i < m; ++i) {
    int type, l, r, k;
    scanf("%d", &type);
    if (type == 1) {
      scanf("%d%d", &l, &r);
      l = (l + lastans - 1) % n + 1;
      r = (r + lastans - 1) % n + 1;
      if (l > r) swap(l, r);
      if (l == r) continue;
      l--;
      r--;
      int val = del(r);
      add(l, val);
    } else {
      scanf("%d%d%d", &l, &r, &k);
      l = (l + lastans - 1) % n + 1;
      r = (r + lastans - 1) % n + 1;
      if (l > r) swap(l, r);
      k = (k + lastans - 1) % n + 1;
      l--;
      r--;
      int ans = get(l, r, k);
      printf("%d\n", ans);
      lastans = ans;
    }
    if (i % 400 == 0) build(e);
  }
  return 0;
}
