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;
int Pow(int a, int b, int md, int ans = 1) {
  for (; b; b >>= 1, a = a * a % md)
    if (b & 1) ans = ans * a % md;
  return ans % md;
}
const int MAXN = 1e5 + 10;
const int MOD = 1e9 + 7;
const int SQ = 512;
const int tof = 511;
template <class T, int SZ>
struct FastDeque {
  T Q[SZ * 2 + 10];
  int l = SZ, r = SZ;
  void push_back(T x) { Q[r++] = x; }
  void push_front(T x) { Q[--l] = x; }
  void pop_front() { l++; }
  void pop_back() { r--; }
  T front() { return Q[l]; }
  T back() { return Q[r - 1]; }
  void clear() { l = r = SZ; }
  bool empty() { return (l == r); }
  T get(int id) { return Q[l + id]; }
};
deque<int> dq[SQ];
int cnt[SQ][MAXN], A[MAXN], n, q, lst;
void updsame(int l, int r) {
  int x = dq[l / SQ][r & tof];
  for (int i = r & tof; i > (l & tof); i--) {
    dq[l / SQ][i] = dq[l / SQ][i - 1];
  }
  dq[l / SQ][l & tof] = x;
}
void updnot(int l, int r) {
  for (int i = l - (l & tof) + SQ - 1; i < r; i += SQ) {
    int x = dq[i / SQ].back();
    dq[i / SQ].pop_back();
    dq[i / SQ + 1].push_front(x);
    cnt[i / SQ][x]--;
    cnt[i / SQ + 1][x]++;
  }
  int x = dq[r / SQ][(r & tof) + 1];
  for (int i = (r & tof); i >= 0; i--) dq[r / SQ][i + 1] = dq[r / SQ][i];
  dq[r / SQ].pop_front();
  dq[l / SQ].push_back(x);
  for (int i = SQ - 1; i > (l & tof); i--) dq[l / SQ][i] = dq[l / SQ][i - 1];
  cnt[r / SQ][x]--, cnt[l / SQ][x]++;
  dq[l / SQ][l & tof] = x;
}
void update(int l, int r) {
  if (l == r) return;
  if (r / SQ == l / SQ)
    updsame(l, r);
  else
    updnot(l, r);
}
int get(int l, int r, int x, int ans = 0) {
  for (int i = l; i <= r;) {
    if ((i & tof) == 0 && i + SQ - 1 <= r) {
      ans += cnt[i / SQ][x];
      i += SQ;
    } else {
      if (dq[i / SQ][i & tof] == x) ans++;
      i++;
    }
  }
  return ans;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &A[i]);
    cnt[i / SQ][A[i]]++;
    dq[i / SQ].push_back(A[i]);
  }
  scanf("%d", &q);
  while (q--) {
    int l, r, t, x;
    scanf("%d%d%d", &t, &l, &r);
    l = (l + lst - 1) % n + 1;
    r = (r + lst - 1) % n + 1;
    if (r < l) swap(l, r);
    l--, r--;
    if (t == 1) {
      update(l, r);
    } else {
      scanf("%d", &x);
      x = (x + lst - 1) % n + 1;
      lst = get(l, r, x);
      printf("%d\n", lst);
    }
  }
  return 0;
}
