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 = 1e5 + 20, K = 19;
const int mod = 1e9 + 7;
int freq[320][N];
int arr[N], s[N];
deque<int> dq[320];
int n, m, len, ans = 0;
void build() {
  for (int i = 0; i < n; i++) {
    freq[i / len][arr[i]]++;
    dq[i / len].push_back(arr[i]);
  }
}
void pr() {
  for (int i = 0; i < n; i++) {
    cout << dq[i / len][i % len] << " ";
  }
  cout << endl;
}
void upd(int l, int r) {
  int lb = l / len;
  int rb = r / len;
  if (lb == rb) {
    int save = dq[r / len][r % len];
    for (int i = r % len; i >= l % len + 1; i--) {
      dq[lb][i] = dq[lb][i - 1];
    }
    dq[lb][l % len] = save;
  } else {
    int lsave = dq[lb].back();
    int rsave = dq[rb][r % len];
    for (int i = len - 1; i >= l % len + 1; i--) {
      dq[lb][i] = dq[lb][i - 1];
    }
    dq[lb][l % len] = rsave;
    freq[lb][rsave]++;
    freq[lb][lsave]--;
    freq[rb][rsave]--;
    for (int i = lb + 1; i < rb; i++) {
      rsave = lsave;
      lsave = dq[i].back();
      dq[i].pop_back();
      dq[i].push_front(rsave);
      freq[i][rsave]++;
      freq[i][lsave]--;
    }
    freq[rb][lsave]++;
    for (int i = r % len; i >= 1; i--) {
      dq[rb][i] = dq[rb][i - 1];
    }
    dq[rb][0] = lsave;
  }
}
void qu(int l, int r, int x) {
  ans = 0;
  int s = l / len;
  int e = r / len;
  for (int i = s + 1; i < e; i++) ans += freq[i][x];
  if (s == e) {
    for (int i = l % len; i <= r % len; i++)
      if (dq[s][i] == x) ans++;
  } else {
    for (int i = l % len; i < len; i++) {
      if (dq[s][i] == x) ans++;
    }
    for (int i = 0; i <= r % len; i++) {
      if (dq[e][i] == x) ans++;
    }
  }
  printf("%d\n", ans);
}
int main() {
  cin >> n;
  len = ceil(sqrt(n));
  for (int i = 0; i < n; i++) scanf("%d", arr + i);
  build();
  cin >> m;
  for (int i = 0; i < m; i++) {
    int ty;
    scanf("%d", &ty);
    if (ty == 1) {
      int l, r;
      scanf("%d%d", &l, &r);
      l = (l + ans - 1) % n;
      r = (r + ans - 1) % n;
      if (r < l) swap(l, r);
      upd(l, r);
    } else {
      int l, r, x;
      scanf("%d%d%d", &l, &r, &x);
      l = (l + ans - 1) % n;
      r = (r + ans - 1) % n;
      if (r < l) swap(l, r);
      x = (x + ans - 1) % n + 1;
      qu(l, r, x);
    }
  }
  return 0;
}