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>
const int MAXN = 1e5 + 10, jazr = 500;
using namespace std;
int unq[jazr][MAXN];
deque<int> dq[500];
vector<int> help;
void inseg(int l, int r) {
  int sgl = l / jazr, sgr = r / jazr, L = l % jazr, R = r % jazr,
      sz = dq[sgl].size();
  for (int i = 1; i < sz - R; i++) {
    help.push_back(dq[sgl].back());
    dq[sgl].pop_back();
  }
  int x = dq[sgr].back();
  dq[sgl].pop_back();
  for (int i = 1; i <= R - L; i++) {
    help.push_back(dq[sgl].back());
    dq[sgl].pop_back();
  }
  dq[sgl].push_back(x);
  while (help.size()) {
    dq[sgl].push_back(help.back());
    help.pop_back();
  }
}
void outseg(int l, int r) {
  int sgl = l / jazr, sgr = r / jazr, L = l % jazr, R = r % jazr,
      sz = dq[sgl].size(), j;
  for (int i = 0; i < R; i++) {
    help.push_back(dq[sgr].front());
    dq[sgr].pop_front();
  }
  j = dq[sgr].front();
  dq[sgr].pop_front();
  unq[sgr][j]--;
  while (help.size()) {
    dq[sgr].push_front(help.back());
    help.pop_back();
  }
  for (int i = sgr; i > sgl; i--) {
    int x = dq[i - 1].back();
    dq[i].push_front(x);
    dq[i - 1].pop_back();
    unq[i][x]++;
    unq[i - 1][x]--;
  }
  sz--;
  for (int i = 1; i <= sz - L; i++) {
    help.push_back(dq[sgl].back());
    dq[sgl].pop_back();
  }
  dq[sgl].push_back(j);
  unq[sgl][j]++;
  while (help.size()) {
    dq[sgl].push_back(help.back());
    help.pop_back();
  }
}
void shift(int l, int r) {
  if (l / jazr == r / jazr)
    inseg(l, r);
  else
    outseg(l, r);
}
int get(int l, int r, int k) {
  int sgl = l / jazr, sgr = r / jazr, L = l % jazr, R = r % jazr,
      sz = dq[sgl].size(), sum = 0;
  if (sgl == sgr) {
    for (int i = 1; i < sz - R; i++) {
      help.push_back(dq[sgl].back());
      dq[sgl].pop_back();
    }
    for (int i = 1; i <= R - L + 1; i++) {
      if (dq[sgl].back() == k) sum++;
      help.push_back(dq[sgl].back());
      dq[sgl].pop_back();
    }
    while (help.size()) {
      dq[sgl].push_back(help.back());
      help.pop_back();
    }
    return sum;
  } else {
    for (int i = 1; i <= sz - L; i++) {
      if (dq[sgl].back() == k) sum++;
      help.push_back(dq[sgl].back());
      dq[sgl].pop_back();
    }
    while (help.size()) {
      dq[sgl].push_back(help.back());
      help.pop_back();
    }
    for (int i = sgl + 1; i < sgr; i++) sum += unq[i][k];
    for (int i = 0; i <= R; i++) {
      if (dq[sgr].front() == k) sum++;
      help.push_back(dq[sgr].front());
      dq[sgr].pop_front();
    }
    while (help.size()) {
      dq[sgr].push_front(help.back());
      help.pop_back();
    }
    return sum;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie();
  cout.tie();
  int n, cnt = -1;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    if (i % jazr == 0) cnt++;
    dq[cnt].push_back(a);
    unq[cnt][a]++;
  }
  int q;
  cin >> q;
  int la = 0;
  while (q--) {
    int c, l, r, k;
    cin >> c;
    if (c == 1) {
      cin >> l >> r;
      l = (l + la - 1) % n + 1;
      r = (r + la - 1) % n + 1;
      if (l > r) swap(l, r);
      l--;
      r--;
      shift(l, r);
    } else {
      cin >> l >> r >> k;
      l = (l + la - 1) % n + 1;
      r = (r + la - 1) % n + 1;
      k = (k + la - 1) % n + 1;
      if (l > r) swap(l, r);
      l--;
      r--;
      la = get(l, r, k);
      cout << la << endl;
    }
  }
}