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 int N = 1e5;
const int B = 350;
int a[N];
int b[N];
int tot = 0;
mt19937 rng(4112003);
struct Node {
  Node *l, *r;
  Node *p;
  int key, pri;
  int nCh, cnt[B];
  Node(int x = 0) : key(x), pri(rng() << 16 | rng()), nCh(1) {
    l = r = 0;
    for (int i = 0; i < tot; ++i) cnt[i] = 0;
    if (b[x] >= 0) cnt[b[x]] = 1;
  };
} * p[N];
int Siz(Node *x) {
  if (x)
    return x->nCh;
  else
    return 0;
}
void pull(Node *x) {
  Node *u = x->l;
  Node *v = x->r;
  x->nCh = 1;
  x->nCh += Siz(u);
  x->nCh += Siz(v);
  for (int i = 0; i < tot; ++i) {
    x->cnt[i] = 0;
    if (u) x->cnt[i] += u->cnt[i];
    if (v) x->cnt[i] += v->cnt[i];
  }
  if (b[x->key] >= 0) x->cnt[b[x->key]]++;
  if (u) u->p = x;
  if (v) v->p = x;
}
Node *join(Node *x, Node *y) {
  if (!x) return y;
  if (!y) return x;
  if (x->pri > y->pri) {
    x->r = join(x->r, y);
    pull(x);
    return x;
  } else {
    y->l = join(x, y->l);
    pull(y);
    return y;
  }
}
void split(Node *x, Node *&l, Node *&r, int p) {
  if (!x) {
    l = r = 0;
    return;
  }
  if (p >= Siz(x->l) + 1) {
    p -= Siz(x->l) + 1;
    split(x->r, x->r, r, p);
    pull(l = x);
  } else
    split(x->l, l, x->l, p), pull(r = x);
}
vector<int> pos[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  srand(time(NULL));
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    int x;
    cin >> x;
    --x;
    pos[x].emplace_back(i);
    a[i] = x;
  }
  for (int i = 0; i < n; ++i) {
    if (pos[i].size() > B)
      b[i] = tot++;
    else
      b[i] = -1;
  }
  Node *R = 0;
  for (int i = 0; i < n; ++i) R = join(R, p[i] = new Node(a[i]));
  int q;
  cin >> q;
  int lst = 0;
  for (int i = 0; i < q; ++i) {
    int t;
    cin >> t;
    int l;
    cin >> l;
    l = (l + lst - 1) % n;
    int r;
    cin >> r;
    r = (r + lst - 1) % n;
    if (l > r) swap(l, r);
    if (t == 1) {
      Node *x, *y, *z;
      split(R, R, x, l);
      split(x, x, y, r - l + 1);
      split(x, x, z, r - l);
      x = join(z, x);
      x = join(x, y);
      R = join(R, x);
    }
    if (t == 2) {
      int x;
      cin >> x;
      x = (x + lst - 1) % n;
      lst = 0;
      if (b[x] >= 0) {
        Node *p, *q;
        split(R, R, p, l);
        split(p, p, q, r - l + 1);
        lst = p->cnt[b[x]];
        p = join(p, q);
        R = join(R, p);
      } else {
        for (int a : pos[x]) {
          Node *T = p[a];
          int cur = Siz(T->l);
          while (T != R) {
            if (T->p->r == T) cur += Siz(T->p->l) + 1;
            T = T->p;
          }
          if (cur < l) continue;
          if (cur > r) continue;
          lst++;
        }
      }
      cout << lst << "\n";
    }
  }
}