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 n, ar[200000];
deque<long> block[1000];
long calc[300][100100];
long tests, tp, l, r, lastans;
long temp;
long k;
long bnum, qpos;
deque<long>::iterator it;
long val;
long rec;
long gett(long ps) {
  long c = ps / 400;
  ps %= 400;
  it = block[c].begin();
  for (int i = 0; i < ps; i++) ++it;
  return (*it);
}
long gett(long l, long r, long val) {
  long res = 0;
  if (l / 400 == r / 400) {
    bnum = l / 400;
    qpos = bnum * 400;
    for (it = block[bnum].begin(); it != block[bnum].end(); it++) {
      if (qpos >= l && qpos <= r && ((*it) == val)) ++res;
      ++qpos;
    }
    return res;
  }
  bnum = l / 400;
  qpos = bnum * 400;
  for (it = block[bnum].begin(); it != block[bnum].end(); it++) {
    if (qpos >= l && qpos <= r && ((*it) == val)) ++res;
    ++qpos;
  }
  for (int i = l / 400 + 1; i < r / 400; i++) res += calc[i][val];
  bnum = r / 400;
  qpos = bnum * 400;
  for (it = block[bnum].begin(); it != block[bnum].end(); it++) {
    if (qpos >= l && qpos <= r && ((*it) == val)) ++res;
    ++qpos;
  }
  return res;
}
void rebuild(long b, long l, long r) {
  long c = 0;
  for (it = block[b].begin(); it != block[b].end(); it++) {
    ar[c] = (*it);
    ++c;
  }
  long temp = ar[r];
  for (int i = r; i > l; i--) ar[i] = ar[i - 1];
  ar[l] = temp;
  while (block[b].size()) block[b].pop_back();
  for (int i = 0; i < c; i++) block[b].push_back(ar[i]);
}
void update(long l, long r) {
  if (l / 400 == r / 400) {
    rebuild(l / 400, l % 400, r % 400);
    return;
  }
  val = gett(r);
  bnum = l / 400;
  long c = 0;
  for (it = block[bnum].begin(); it != block[bnum].end(); it++) {
    ar[c] = (*it);
    ++c;
  }
  for (int i = c; i > l % 400; i--) ar[i] = ar[i - 1];
  ar[l % 400] = val;
  while (block[bnum].size()) {
    temp = block[bnum].back();
    calc[bnum][temp]--;
    block[bnum].pop_back();
  }
  for (int i = 0; i < c; i++) {
    block[bnum].push_back(ar[i]);
    calc[bnum][ar[i]]++;
  }
  rec = ar[c];
  for (int i = l / 400 + 1; i < r / 400; i++) {
    calc[i][rec]++;
    block[i].push_front(rec);
    rec = block[i].back();
    calc[i][rec]--;
    block[i].pop_back();
  }
  bnum = r / 400;
  c = 0;
  for (it = block[bnum].begin(); it != block[bnum].end(); it++) {
    ar[c] = (*it);
    ++c;
  }
  for (int i = r % 400; i > 0; i--) ar[i] = ar[i - 1];
  ar[0] = rec;
  while (block[bnum].size()) {
    temp = block[bnum].back();
    calc[bnum][temp]--;
    block[bnum].pop_back();
  }
  for (int i = 0; i < c; i++) {
    block[bnum].push_back(ar[i]);
    calc[bnum][ar[i]]++;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> ar[i];
  for (int i = 0; i < n; i++) {
    block[i / 400].push_back(ar[i]);
    calc[i / 400][ar[i]]++;
  }
  cin >> tests;
  for (; tests; --tests) {
    cin >> tp;
    cin >> l >> r;
    l = l + lastans - 1;
    l = l % n + 1;
    r = r + lastans - 1;
    r = r % n + 1;
    if (r < l) swap(r, l);
    if (tp == 2) cin >> k;
    k = k + lastans - 1;
    k = k % n + 1;
    if (tp == 2) {
      temp = gett(l - 1, r - 1, k);
      cout << temp << endl;
      lastans = temp;
      continue;
    } else {
      update(l - 1, r - 1);
    }
  }
  cin.get();
  cin.get();
  return 0;
}