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 SZ = 333;
const int MAX = 100010;
struct block {
  int a[3 * SZ];
  int st, fn;
  int* col;
  block(const vector<int>& b, int* mas) {
    st = SZ;
    fn = int(b.size()) + SZ;
    col = mas;
    for (int i = st; i < fn; ++i) {
      a[i] = b[i - st];
      col[a[i]]++;
    }
  }
  void del(int idx) {
    col[a[st + idx]]--;
    fn--;
    for (int i = st + idx; i < fn; ++i) {
      a[i] = a[i + 1];
    }
  }
  void insert(int b, int idx) {
    col[b]++;
    for (int i = fn; i > st + idx; --i) {
      a[i] = a[i - 1];
    }
    a[st + idx] = b;
    fn++;
  }
  inline int& operator[](int idx) { return a[idx + st]; }
  inline int size() { return fn - st; }
};
int col[MAX / SZ + 1][MAX];
int n;
int a[100100];
vector<block> w;
void build() {
  w.clear();
  for (int i = 0, cnt = 0; i < n; i += SZ, cnt++) {
    int r = min(n, i + SZ);
    vector<int> help;
    for (int j = i; j < r; ++j) {
      help.push_back(a[j]);
    }
    block loc(help, col[cnt]);
    w.push_back(loc);
  }
}
void rebuild() {
  int idx = 0;
  for (int i = 0; i < int(w.size()); ++i) {
    for (int j = 0; j < int(w[i].size()); ++j) {
      a[idx++] = w[i][j];
      w[i].col[w[i][j]]--;
    }
  }
  build();
}
inline int fnd(int l, int r, int k) {
  int tek = 0;
  int ans = 0;
  for (int i = 0; i < int(w.size()); ++i) {
    int col = w[i].size();
    if (tek + col < l) {
      tek += col;
      continue;
    }
    if (tek >= r) {
      break;
    }
    if (tek + col <= r && tek + 1 >= l) {
      ans += w[i].col[k];
      tek += col;
      continue;
    }
    int l1 = max(l - tek - 1, 0), r1 = min(col - 1, r - tek - 1);
    for (int j = l1; j <= r1; ++j) {
      ans += (w[i][j] == k);
    }
    tek += col;
  }
  return ans;
}
inline void upd(int l, int r) {
  int tek = 0;
  int elem = -1;
  for (int i = 0; i < int(w.size()); ++i) {
    int col = w[i].size();
    if (tek + col < r) {
      tek += col;
      continue;
    }
    int idx = r - tek - 1;
    elem = w[i][idx];
    w[i].del(idx);
    break;
  }
  tek = 0;
  for (int i = 0; i < int(w.size()); ++i) {
    int col = w[i].size();
    if (tek + col + 1 < l) {
      tek += col;
      continue;
    }
    int idx = l - tek - 1;
    w[i].insert(elem, idx);
    break;
  }
}
int main() {
  cin >> n;
  for (int i = 0; i < int(n); ++i) {
    cin >> a[i];
  }
  build();
  int q;
  cin >> q;
  int col_without = 0;
  int lastans = 0;
  for (int i = 0; i < int(q); ++i) {
    int tp;
    scanf("%d", &tp);
    if (tp == 1) {
      col_without++;
      int l, r;
      scanf("%d %d", &l, &r);
      l = ((l + lastans - 1) % n) + 1;
      r = ((r + lastans - 1) % n) + 1;
      if (l > r) swap(l, r);
      upd(l, r);
    } else {
      int l, r, k;
      scanf("%d %d %d", &l, &r, &k);
      l = ((l + lastans - 1) % n) + 1;
      r = ((r + lastans - 1) % n) + 1;
      k = ((k + lastans - 1) % n) + 1;
      if (l > r) swap(l, r);
      printf("%d\n", lastans = fnd(l, r, k));
    }
    if (col_without >= SZ) {
      rebuild();
      col_without = 0;
    }
  }
  return 0;
}