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
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 318;
const int INF = 0x7f7f7f7f;
int cases, len, ans, lastans;
template <class T>
inline void read(T& x) {
  int c;
  for (c = getchar(); c <= 32 && ~c; c = getchar())
    ;
  for (x = 0; c > 32; x = x * 10 + c - '0', c = getchar())
    ;
}
deque<int> block[MAXN];
int cnt[MAXN][100001];
void shift(int l, int r) {
  int st = l / len, ed = r / len;
  int posl = l % len, posr = r % len;
  if (st == ed) {
    int tmp = block[st][posr];
    for (int i = posr; i > posl; i--) {
      block[st][i] = block[st][i - 1];
    }
    block[st][posl] = tmp;
  } else {
    int tmp = block[ed][posr];
    cnt[ed][tmp]--;
    block[st + 1].push_front(block[st].back());
    cnt[st][block[st].back()]--;
    cnt[st + 1][block[st].back()]++;
    for (int i = len - 1; i > posl; i--) {
      block[st][i] = block[st][i - 1];
    }
    block[st][posl] = tmp;
    cnt[st][tmp]++;
    for (int i = st + 1; i < ed; i++) {
      cnt[i][block[i].back()]--;
      cnt[i + 1][block[i].back()]++;
      block[i + 1].push_front(block[i].back());
      block[i].pop_back();
    }
    for (int i = posr + 1; i > 0; i--) {
      block[ed][i] = block[ed][i - 1];
    }
    block[ed].pop_front();
  }
}
int query(int l, int r, int k) {
  int st = l / len, ed = r / len;
  int posl = l % len, posr = r % len;
  int ret = 0;
  if (st == ed) {
    for (int i = posl; i <= posr; i++) {
      if (block[st][i] == k) ++ret;
    }
  } else {
    for (int i = st + 1; i <= ed - 1; i++) {
      ret += cnt[i][k];
    }
    for (int i = posl; i < len; i++) {
      if (block[st][i] == k) ++ret;
    }
    for (int i = 0; i <= posr; i++) {
      if (block[ed][i] == k) ++ret;
    }
  }
  return ret;
}
int main() {
  int n, q, x, y, k, op, tot = -1;
  read(n);
  len = sqrt(n);
  for (int i = 0; i < n; i++) {
    read(x);
    if (i % len == 0) {
      tot++;
    }
    block[tot].push_back(x);
    cnt[tot][x]++;
  }
  read(q);
  while (q--) {
    read(op);
    if (op == 1) {
      read(x), read(y);
      x = (x + lastans - 1) % n + 1;
      y = (y + lastans - 1) % n + 1;
      x--, y--;
      if (x > y) swap(x, y);
      shift(x, y);
    } else {
      read(x), read(y), read(k);
      x = (x + lastans - 1) % n + 1;
      y = (y + lastans - 1) % n + 1;
      k = (k + lastans - 1) % n + 1;
      x--, y--;
      if (x > y) swap(x, y);
      lastans = query(x, y, k);
      printf("%d\n", lastans);
    }
  }
}
