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 N = 100005, BLOCK_SIZE = 320;
int n, a[N], rep[BLOCK_SIZE][N];
deque<int> blocks[BLOCK_SIZE];
void update(int l, int r) {
  int lBlock = l / BLOCK_SIZE, rBlock = r / BLOCK_SIZE;
  int cur = blocks[rBlock][r % BLOCK_SIZE];
  blocks[rBlock].erase(blocks[rBlock].begin() + r % BLOCK_SIZE);
  --rep[rBlock][cur];
  blocks[lBlock].insert(blocks[lBlock].begin() + l % BLOCK_SIZE, cur);
  ++rep[lBlock][cur];
  if ((int)blocks[lBlock].size() > BLOCK_SIZE)
    cur = blocks[lBlock].back(), blocks[lBlock].pop_back(), --rep[lBlock][cur];
  for (int(i) = (lBlock + 1); (i) <= (int)(rBlock); (i)++) {
    blocks[i].push_front(cur);
    ++rep[i][cur];
    if ((int)blocks[i].size() > BLOCK_SIZE) {
      cur = blocks[i].back(), blocks[i].pop_back();
      --rep[i][cur];
    }
  }
}
int query(int l, int r, int k) {
  int ans = 0;
  int lBlock = l / BLOCK_SIZE, rBlock = r / BLOCK_SIZE;
  while (l <= r && l % BLOCK_SIZE) {
    if (blocks[lBlock][l % BLOCK_SIZE] == k) ++ans;
    ++l;
  }
  while (r >= l && r % BLOCK_SIZE != BLOCK_SIZE - 1 && r >= 0) {
    if (blocks[rBlock][r % BLOCK_SIZE] == k) ++ans;
    --r;
  }
  if (l < r) {
    lBlock = l / BLOCK_SIZE, rBlock = r / BLOCK_SIZE;
    for (int(i) = (lBlock); (i) <= (int)(rBlock); (i)++) ans += rep[i][k];
  }
  return ans;
}
int main() {
  scanf("%d", &n);
  int blockNum = 0;
  for (int(i) = (0); (i) <= (int)(n - 1); (i)++) {
    scanf("%d", &a[i]);
    blocks[blockNum].push_back(a[i]);
    ++rep[blockNum][a[i]];
    if ((int)blocks[blockNum].size() == BLOCK_SIZE) ++blockNum;
  }
  int q, ans = 0;
  scanf("%d", &q);
  while (q--) {
    int t, l, r;
    scanf("%d%d%d", &t, &l, &r);
    l = (l + ans - 1) % n, r = (r + ans - 1) % n;
    if (l > r) swap(l, r);
    if (1 == t)
      update(l, r);
    else {
      int k;
      scanf("%d", &k);
      k = (k + ans - 1) % n + 1;
      printf("%d\n", ans = query(l, r, k));
    }
  }
  return 0;
}
