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 long long INF = (long long)1e9 + 5;
const int MAXN = 100005;
const int SZ = 300;
const int BLK = MAXN / SZ + 5;
int N, B, a[MAXN], lastans;
deque<int> deq[BLK];
int cnt[BLK][MAXN];
void build() {
  B = (N + SZ - 1) / SZ;
  for (int i = 0; i < B; i++) {
    for (int j = i * SZ; j < (i + 1) * SZ; j++) {
      deq[i].push_back(a[j]);
      cnt[i][a[j]]++;
    }
  }
}
void do_shift(int l, int r) {
  int lb = l / SZ, rb = r / SZ;
  int st_l = lb * SZ, st_r = rb * SZ;
  if (lb == rb) {
    l -= st_l, r -= st_l;
    vector<int> tmp(deq[lb].begin(), deq[lb].end());
    deq[lb].clear();
    for (int i = 0; i < l; i++) {
      deq[lb].push_back(tmp[i]);
    }
    deq[lb].push_back(tmp[r]);
    for (int i = l; i < r; i++) {
      deq[lb].push_back(tmp[i]);
    }
    for (int i = r + 1; i < SZ; i++) {
      deq[lb].push_back(tmp[i]);
    }
    return;
  }
  auto it1 = deq[lb].begin(), it2 = deq[rb].begin();
  for (int i = 0; i < l - st_l; i++) ++it1;
  for (int i = 0; i < r - st_r; i++) ++it2;
  cnt[lb][*it2]++;
  deq[lb].insert(it1, *it2);
  cnt[rb][*it2]--;
  deq[rb].erase(it2);
  for (int i = lb + 1; i <= rb; i++) {
    cnt[i][deq[i - 1].back()]++;
    cnt[i - 1][deq[i - 1].back()]--;
    deq[i].push_front(deq[i - 1].back());
    deq[i - 1].pop_back();
  }
}
int do_count(int l, int r, int k) {
  int lb = l / SZ, rb = r / SZ, res = 0;
  if (lb == rb) {
    auto it = deq[lb].begin();
    for (int i = 0; i < l - lb * SZ; i++) ++it;
    for (int i = 0; i < r - l + 1; i++) {
      res += (*it == k);
      ++it;
    }
    return res;
  }
  auto it = deq[lb].begin();
  for (int i = 0; i < l - lb * SZ; i++) ++it;
  while (it != deq[lb].end()) {
    res += (*it == k);
    ++it;
  }
  it = deq[rb].begin();
  for (int i = 0; i < r - rb * SZ + 1; i++) {
    res += (*it == k);
    ++it;
  }
  for (int i = lb + 1; i < rb; i++) {
    res += cnt[i][k];
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> N;
  int mod = N;
  for (int i = 0; i < N; i++) {
    cin >> a[i];
  }
  while (N % SZ != 0) {
    a[N++] = 0;
  }
  build();
  int Q;
  cin >> Q;
  for (int i = 0; i < Q; i++) {
    int t, l, r, k;
    cin >> t >> l >> r;
    if (t == 2) cin >> k;
    l = (l + lastans - 1) % mod;
    r = (r + lastans - 1) % mod;
    k = (k + lastans - 1) % mod + 1;
    if (l > r) swap(l, r);
    if (t == 1) {
      do_shift(l, r);
    } else {
      int res = do_count(l, r, k);
      cout << res << '\n';
      lastans = res;
    }
  }
  return 0;
}