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;
long long int arr[100005], has[317][100005], value[317][3 * 317], siz[317], n,
    cur;
void build() {
  for (long long int i = 1; i <= n; i += 317) {
    long long int x = min(n, i + 317 - 1);
    siz[cur] = x - i + 1;
    long long int z = 1;
    for (long long int j = i; j <= x; j++) {
      has[cur][arr[j]]++;
      value[cur][z] = arr[j];
      z++;
    }
    cur++;
  }
}
void rebuild() {
  long long int x = 1;
  for (long long int i = 0; i < 317; i++) {
    for (long long int j = 1; j <= siz[i]; j++) {
      arr[x] = value[i][j];
      has[i][value[i][j]]--;
      x++;
    }
  }
  memset(siz, 0, sizeof(siz));
  cur = 0;
  build();
}
long long int remove(long long int x) {
  long long int curr = 0, res = 0;
  for (long long int i = 0; i < 317; i++) {
    curr += siz[i];
    if (curr >= x) {
      long long int index = siz[i] - (curr - x);
      has[i][value[i][index]]--;
      res = value[i][index];
      for (long long int j = index; j < siz[i]; j++)
        value[i][j] = value[i][j + 1];
      siz[i]--;
      break;
    }
  }
  return res;
}
void insert(long long int x, long long int val) {
  long long int curr = 0;
  for (long long int i = 0; i < 317; i++) {
    curr += siz[i];
    if (curr >= x) {
      long long int index = siz[i] - (curr - x);
      has[i][val]++;
      siz[i]++;
      for (long long int j = siz[i]; j > index + 1; j--)
        value[i][j] = value[i][j - 1];
      value[i][index + 1] = val;
      break;
    }
  }
}
void update(long long int l, long long int r) {
  if (l == r) return;
  long long int val = remove(r);
  insert(l - 1, val);
}
long long int query(long long int l, long long int r, long long int k) {
  long long int curr = 0, res = 0;
  for (long long int i = 0; i < 317; i++) {
    curr += siz[i];
    if (curr >= l && curr >= r) {
      long long int index1 = siz[i] - (curr - l);
      long long int index2 = siz[i] - (curr - r);
      for (long long int j = index1; j <= index2; j++) {
        if (value[i][j] == k) res++;
      }
      return res;
    }
    if (curr >= l) {
      long long int index1 = siz[i] - (curr - l);
      for (long long int j = index1; j <= siz[i]; j++) {
        if (value[i][j] == k) res++;
      }
      for (long long int j = i + 1; j < 317; j++) {
        curr += siz[j];
        if (curr >= r) {
          long long int index2 = siz[j] - (curr - r);
          for (long long int t = 1; t <= index2; t++) {
            if (value[j][t] == k) res++;
          }
          return res;
        } else
          res += has[j][k];
      }
    }
  }
  return res;
}
int main() {
  scanf("%lld", &n);
  for (long long int i = 1; i <= n; i++) scanf("%lld", &arr[i]);
  build();
  long long int q;
  scanf("%lld", &q);
  long long int last = 0, z = 1;
  for (long long int i = 0; i < q; i++) {
    long long int a, l, r, k;
    scanf("%lld", &a);
    if (a == 1) {
      scanf("%lld", &l);
      scanf("%lld", &r);
      l = (l + last - 1) % n + 1;
      r = (r + last - 1) % n + 1;
      if (l > r) swap(l, r);
      update(l, r);
      z++;
    } else {
      scanf("%lld", &l);
      scanf("%lld", &r);
      scanf("%lld", &k);
      l = (l + last - 1) % n + 1;
      r = (r + last - 1) % n + 1;
      k = (k + last - 1) % n + 1;
      if (l > r) swap(l, r);
      last = query(l, r, k);
      printf("%lld\n", last);
    }
    if (z % 317 == 0) {
      rebuild();
      z++;
    }
  }
  return 0;
}