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 = 200010;
const int MAGIC = 474;
int len;
int a[N];
int cnt, from[N], to[N];
vector<int> where[N];
void init() {
  cnt = 1;
  from[0] = 1;
  to[0] = len;
  for (int i = 1; i <= len; i++) {
    where[i].clear();
  }
  for (int i = 1; i <= len; i++) {
    where[a[i]].push_back(i);
  }
}
int get_count(int k, int from, int to) {
  int sz = where[k].size();
  if (sz == 0) {
    return 0;
  }
  int low = 0, high = sz;
  while (low < high) {
    int mid = (low + high) >> 1;
    if (where[k][mid] < from) {
      low = mid + 1;
    } else {
      high = mid;
    }
  }
  int X = low;
  low = -1, high = sz - 1;
  while (low < high) {
    int mid = (low + high + 1) >> 1;
    if (where[k][mid] > to) {
      high = mid - 1;
    } else {
      low = mid;
    }
  }
  int Y = low;
  return (Y - X + 1);
}
int new_from[N], new_to[N];
int bounds[4], bound_id;
void cut(int b0, int b1, int b2) {
  bounds[0] = b0;
  bounds[1] = b1;
  bounds[2] = b2;
  bounds[3] = 123456789;
  bound_id = 0;
  int new_cnt = 0;
  int sum = 0;
  for (int i = 0; i < cnt; i++) {
    int new_sum = sum + (to[i] - from[i] + 1);
    if (new_sum <= bounds[bound_id]) {
      if (new_sum == bounds[bound_id]) {
        bound_id++;
      }
      new_from[new_cnt] = from[i];
      new_to[new_cnt] = to[i];
      new_cnt++;
      sum = new_sum;
      continue;
    }
    int get = bounds[bound_id] - sum;
    new_from[new_cnt] = from[i];
    new_to[new_cnt] = from[i] + get - 1;
    new_cnt++;
    from[i] += get;
    sum += get;
    bound_id++;
    i--;
  }
  cnt = new_cnt;
  for (int i = 0; i < cnt; i++) {
    from[i] = new_from[i];
    to[i] = new_to[i];
  }
}
int new_a[N];
int main() {
  scanf("%d", &len);
  for (int i = 1; i <= len; i++) {
    scanf("%d", a + i);
  }
  init();
  int tt;
  scanf("%d", &tt);
  int lastans = 0;
  for (int qq = 1; qq <= tt; qq++) {
    int com;
    scanf("%d", &com);
    if (com == 2) {
      int ll, rr, kk;
      scanf("%d %d %d", &ll, &rr, &kk);
      ll = ((ll + lastans - 1) % len) + 1;
      rr = ((rr + lastans - 1) % len) + 1;
      kk = ((kk + lastans - 1) % len) + 1;
      if (ll > rr) {
        swap(ll, rr);
      }
      int ans = 0;
      int sum = 0;
      for (int i = 0; i < cnt; i++) {
        int x = from[i], y = to[i];
        int begin = sum + 1, end = sum + (y - x + 1);
        if (ll > begin) {
          x += ll - begin;
        }
        if (rr < end) {
          y -= end - rr;
        }
        if (x <= y) {
          ans += get_count(kk, x, y);
        }
        sum = end;
      }
      lastans = ans;
      printf("%d\n", ans);
    } else {
      int ll, rr;
      scanf("%d %d", &ll, &rr);
      ll = ((ll + lastans - 1) % len) + 1;
      rr = ((rr + lastans - 1) % len) + 1;
      if (ll > rr) {
        swap(ll, rr);
      }
      if (ll == rr) {
        continue;
      }
      cut(ll - 1, rr - 1, rr);
      int first = -1;
      int sum = 0;
      for (int i = 0; i < cnt; i++) {
        if (sum == ll - 1) {
          first = i;
        }
        sum += to[i] - from[i] + 1;
        if (sum == rr) {
          int p = from[i];
          for (int j = i; j > first; j--) {
            from[j] = from[j - 1];
            to[j] = to[j - 1];
          }
          from[first] = to[first] = p;
          break;
        }
      }
    }
    if (cnt > MAGIC) {
      int pos = 0;
      for (int i = 0; i < cnt; i++) {
        for (int j = from[i]; j <= to[i]; j++) {
          new_a[++pos] = a[j];
        }
      }
      for (int i = 1; i <= len; i++) {
        a[i] = new_a[i];
      }
      init();
    }
  }
  return 0;
}