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 S = 100003;
int n, a[S], f[353][S], nn, bl[S], fr[353], to[353], m, bg[353], ed[353],
    las = 0;
deque<int> q[353];
void read(int &s) {
  s = 0;
  char c = getchar(), t = 1;
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  if (c == '-') c = getchar(), t = -1;
  while (c >= '0' && c <= '9')
    s = (s << 1) + (s << 3) + (c ^ 48), c = getchar();
  s *= t;
}
char colp = 0;
int main() {
  read(n);
  nn = sqrt(n);
  colp = 'a';
  for (int i = 1; i <= n; ++i) read(a[i]);
  for (int i = 1; i <= n; ++i) bl[i] = (i - 1) / nn + 1;
  for (int i = 1; i <= n; ++i) {
    if (bl[i] != bl[i - 1]) fr[bl[i]] = i;
    q[bl[i]].push_back(a[i]);
    ++f[bl[i]][a[i]];
    if (bl[i] != bl[i + 1]) to[bl[i]] = i;
  }
  read(m);
  int op, l, r, k;
  while (m--) {
    read(op);
    read(l);
    read(r);
    l = (l + las - 1) % n + 1;
    r = (r + las - 1) % n + 1;
    if (l > r) l ^= r ^= l ^= r;
    if (op == 1) {
      bg[0] = ed[0] = 0;
      while (!q[bl[l]].empty())
        bg[++bg[0]] = q[bl[l]].front(), q[bl[l]].pop_front();
      int l1 = l - fr[bl[l]] + 1, l2 = r - fr[bl[r]] + 1;
      if (bl[l] == bl[r]) {
        for (int i = 1; i < l1; ++i) q[bl[l]].push_back(bg[i]);
        q[bl[l]].push_back(bg[l2]);
        for (int i = l1; i < l2; ++i) q[bl[l]].push_back(bg[i]);
        for (int i = l2 + 1; i <= bg[0]; ++i) q[bl[l]].push_back(bg[i]);
      } else {
        while (!q[bl[r]].empty())
          ed[++ed[0]] = q[bl[r]].front(), q[bl[r]].pop_front();
        q[bl[l] + 1].push_front(bg[bg[0]]);
        ++f[bl[l] + 1][bg[bg[0]]];
        --f[bl[l]][bg[bg[0]]];
        for (int i = bl[l] + 1; i < bl[r]; ++i) {
          int o = q[i].back();
          q[i + 1].push_front(o);
          ++f[i + 1][o];
          q[i].pop_back();
          --f[i][o];
        }
        for (int i = 1; i < l2; ++i) q[bl[r]].push_back(ed[i]);
        for (int i = l2 + 1; i <= ed[0]; ++i) q[bl[r]].push_back(ed[i]);
        for (int i = 1; i < l1; ++i) q[bl[l]].push_back(bg[i]);
        q[bl[l]].push_back(ed[l2]);
        --f[bl[r]][ed[l2]];
        ++f[bl[l]][ed[l2]];
        for (int i = l1; i < bg[0]; ++i) q[bl[l]].push_back(bg[i]);
      }
    } else {
      read(k);
      k = (k + las - 1) % n + 1;
      bg[0] = ed[0] = 0;
      while (!q[bl[l]].empty())
        bg[++bg[0]] = q[bl[l]].front(), q[bl[l]].pop_front();
      int l1 = l - fr[bl[l]] + 1, l2 = r - fr[bl[r]] + 1;
      if (bl[l] == bl[r]) {
        int o = 0;
        for (int i = l1; i <= l2; ++i)
          if (bg[i] == k) ++o;
        printf("%d\n", las = o);
      } else {
        int o = 0;
        while (!q[bl[r]].empty())
          ed[++ed[0]] = q[bl[r]].front(), q[bl[r]].pop_front();
        for (int i = bl[l] + 1; i < bl[r]; ++i) o += f[i][k];
        for (int i = l1; i <= bg[0]; ++i)
          if (bg[i] == k) ++o;
        for (int i = 1; i <= l2; ++i)
          if (ed[i] == k) ++o;
        for (int i = 1; i <= ed[0]; ++i) q[bl[r]].push_back(ed[i]);
        printf("%d\n", las = o);
      }
      for (int i = 1; i <= bg[0]; ++i) q[bl[l]].push_back(bg[i]);
    }
  }
  return 0;
}