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;
bool debug = 1;
int n, m, k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
long long ln, lk, lm;
queue<int> vp[100005];
int mp[320][100005];
int a[100005], S;
int getPos(int id, int num) {
  num -= id * S;
  int ans = vp[id].size() - 1 - num;
  return ans;
}
vector<int> change(int id) {
  vector<int> v;
  while (!vp[id].empty()) {
    v.push_back(vp[id].front());
    vp[id].pop();
  }
  return v;
}
void ins(int id, vector<int> &v) {
  assert(vp[id].empty());
  for (int c : v) vp[id].push(c);
}
void shift(int l, int r) {
  int idL = l / S, idR = r / S;
  int posL = getPos(idL, l);
  int posR = getPos(idR, r);
  if (idL == idR) {
    vector<int> v = change(idL);
    int tmp = v[posR];
    for (int i = posR; i < posL; i++) v[i] = v[i + 1];
    v[posL] = tmp;
    ins(idL, v);
    return;
  }
  vector<int> v = change(idR);
  int tmp = v[posR];
  mp[idR][tmp]--;
  for (int i = posR; i < v.size() - 1; i++) {
    v[i] = v[i + 1];
  }
  v.back() = vp[idR - 1].front();
  mp[idR][v.back()]++;
  ins(idR, v);
  for (int b = idR - 1; b > idL; b--) {
    mp[b][vp[b].front()]--;
    vp[b].pop();
    vp[b].push(vp[b - 1].front());
    mp[b][vp[b - 1].front()]++;
  }
  v = change(idL);
  mp[idL][v[0]]--;
  mp[idL][tmp]++;
  for (int i = 0; i < posL; i++) v[i] = v[i + 1];
  v[posL] = tmp;
  ins(idL, v);
}
int qy(int l, int r, int k) {
  int idL = l / S, idR = r / S;
  int posL = getPos(idL, l);
  int posR = getPos(idR, r);
  int ans = 0;
  if (idL == idR) {
    vector<int> v = change(idL);
    for (int i = posR; i <= posL; i++)
      if (v[i] == k) ans++;
    ins(idL, v);
    return ans;
  }
  vector<int> v = change(idL);
  for (int i = 0; i <= posL; i++)
    if (v[i] == k) ans++;
  ins(idL, v);
  for (int i = idL + 1; i < idR; i++) ans += mp[i][k];
  v = change(idR);
  for (int i = posR; i < v.size(); i++)
    if (v[i] == k) ans++;
  ins(idR, v);
  return ans;
}
int main() {
  scanf("%d", &n);
  S = sqrt(n);
  for (int(i) = 0; (i) < (int)(n); (i)++) scanf("%d", a + i);
  for (int i = 0;; i++) {
    int s = i * S;
    int f = min(s + S, n);
    if (s >= f) break;
    for (int j = f - 1; j >= s; j--) {
      vp[i].push(a[j]);
      mp[i][a[j]]++;
    }
  }
  int q, l, r, t, ans = 0;
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d%d", &t, &l, &r);
    l = (l + ans - 1) % n + 1;
    r = (r + ans - 1) % n + 1;
    if (r < l) swap(l, r);
    l--;
    r--;
    if (t == 1) {
      shift(l, r);
    } else {
      scanf("%d", &k);
      k = (k + ans - 1) % n + 1;
      ans = qy(l, r, k);
      printf("%d\n", ans);
    }
  }
  return 0;
}