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 mx = 1e5 + 10;
int a, n;
struct Node {
  deque<int> dq;
  int num[mx];
} bk[100];
int block;
int belong[mx];
int num;
void build() {
  block = pow(mx, 0.618);
  num = (n + block - 1) / block;
  for (int i = 1; i <= n; ++i) belong[i] = (i - 1) / block + 1;
}
void updata(int nl, int nr) {
  int pl = belong[nl], pr = belong[nr];
  if (pl == pr) {
    int st = (pr - 1) * block + 1;
    int p = nl - st;
    int p2 = nr - st;
    int nc = bk[pr].dq[p2];
    bk[pr].dq.erase(bk[pr].dq.begin() + p2);
    bk[pr].dq.insert(bk[pr].dq.begin() + p, nc);
  } else {
    int stl = (pl - 1) * block + 1;
    int enl = pl * block;
    int str = (pr - 1) * block + 1;
    int enr = pr == num ? n : pr * block;
    int p = nl - stl;
    int p2 = nr - str;
    int nc = bk[pr].dq[p2];
    int last = bk[pl].dq.back();
    bk[pl].num[nc]++;
    bk[pl].num[last]--;
    bk[pl].dq.pop_back();
    bk[pl].dq.insert(bk[pl].dq.begin() + p, nc);
    bk[pr].num[nc]--;
    bk[pr].dq.erase(bk[pr].dq.begin() + p2);
    for (int i = pl + 1; i <= pr; ++i) {
      bk[i].dq.push_front(last);
      bk[i].num[last]++;
      last = bk[i].dq.back();
      if (i != pr) {
        bk[i].dq.pop_back();
        bk[i].num[last]--;
      }
    }
  }
}
int query(int l, int r, int c) {
  int ans = 0;
  int pl = belong[l], pr = belong[r];
  if (pl == pr) {
    int st = (pr - 1) * block + 1;
    int en = r - st;
    for (int i = l - st; i <= en; ++i) {
      if (c == bk[pr].dq[i]) ans++;
    }
  } else {
    int stl = (pl - 1) * block + 1;
    int enl = pl * block - stl;
    int str = (pr - 1) * block + 1;
    int enr = r - str;
    for (int i = pl + 1; i <= pr - 1; ++i) {
      ans += bk[i].num[c];
    }
    for (int i = l - stl; i <= enl; ++i) {
      if (c == bk[pl].dq[i]) ans++;
    }
    for (int i = 0; i <= enr; ++i) {
      if (c == bk[pr].dq[i]) ans++;
    }
  }
  return ans;
}
int main() {
  int q, op, l, r, k, lsans = 0;
  scanf("%d", &n);
  build();
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a);
    bk[belong[i]].dq.push_back(a);
    bk[belong[i]].num[a]++;
  }
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d%d", &op, &l, &r);
    l = (l + lsans - 1) % n + 1;
    r = (r + lsans - 1) % n + 1;
    if (l > r) swap(l, r);
    if (op == 1) {
      updata(l, r);
    } else {
      scanf("%d", &k);
      k = (k + lsans - 1) % n + 1;
      lsans = query(l, r, k);
      printf("%d\n", lsans);
    }
  }
  return 0;
}
