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>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int maxn = 1e5 + 10;
int N, bsiz;
list<int> block[350];
unsigned short cnt[350][maxn];
inline void decode(int &x, int last) { x = (x + last - 1) % N + 1; }
void update(int, int);
int query(int, int, int);
int main() {
  bool debug = 0;
  double now_time = clock();
  while (~scanf("%d", &N)) {
    bsiz = sqrt(N) * log2(N);
    for (int i = 0; i <= N / bsiz; i++) block[i].clear();
    memset(cnt, 0, sizeof(cnt));
    for (int i = 0, x; i < N; i++) {
      scanf("%d", &x);
      block[i / bsiz].push_back(x);
      cnt[i / bsiz][x]++;
    }
    int Q, ans = 0;
    scanf("%d", &Q);
    for (int q = 0, t, l, r, k; q < Q; q++) {
      scanf("%d%d%d", &t, &l, &r);
      decode(l, ans), decode(r, ans);
      if (l > r) swap(l, r);
      if (t == 1)
        update(l - 1, r - 1);
      else {
        scanf("%d", &k);
        decode(k, ans);
        ans = query(l - 1, r - 1, k);
        printf("%d\n", ans);
      }
    }
  }
  return 0;
}
void update(int l, int r) {
  if (l == r) return;
  int ll = l / bsiz, rr = r / bsiz, num;
  if (ll == rr) {
    int idxl = ll * bsiz, idxr = (ll + 1) * bsiz - 1;
    if (idxr >= N) idxr = N - 1;
    auto itl = block[ll].begin(), itr = block[ll].end();
    itr--;
    while (idxl < l) {
      idxl++;
      itl++;
    }
    while (idxr > r) {
      idxr--;
      itr--;
    }
    int num = *itr;
    block[ll].erase(itr);
    block[ll].insert(itl, num);
  } else {
    int idx = rr * bsiz;
    auto it = block[rr].begin();
    while (idx < r) {
      idx++;
      it++;
    }
    num = *it;
    block[rr].erase(it);
    cnt[rr][num]--;
    idx = (ll + 1) * bsiz - 1;
    if (idx >= N) idx = N - 1;
    it = block[ll].end();
    it--;
    while (idx > l) {
      idx--;
      it--;
    }
    block[ll].insert(it, num);
    cnt[ll][num]++;
    for (int i = ll; i < rr; i++) {
      num = block[i].back();
      block[i].pop_back();
      block[i + 1].push_front(num);
      cnt[i][num]--, cnt[i + 1][num]++;
    }
  }
}
int query(int l, int r, int k) {
  int ans = 0;
  int ll = l / bsiz, rr = r / bsiz;
  if (ll == rr) {
    int idx = ll * bsiz;
    auto it = block[ll].begin();
    while (idx < l) {
      idx++;
      it++;
    }
    while (idx <= r) {
      if (*it == k) ans++;
      idx++;
      it++;
    }
  } else {
    for (int i = ll + 1; i < rr; i++) ans += cnt[i][k];
    int idx = rr * bsiz;
    auto it = block[rr].begin();
    while (idx <= r) {
      if (*it == k) ans++;
      idx++;
      it++;
    }
    idx = (ll + 1) * bsiz - 1;
    if (idx >= N) idx = N - 1;
    it = block[ll].end();
    it--;
    while (idx >= l) {
      if (*it == k) ans++;
      idx--;
      it--;
    }
  }
  return ans;
}