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 N = 100110;
struct Node;
Node* null;
struct Node {
  Node* c[2];
  Node* fa;
  Node() { c[0] = c[1] = fa = null; }
  Node* link;
  int sz, val;
  inline void setc(int d, Node* s) {
    c[d] = s;
    s->fa = this;
  }
  inline bool d() { return fa->c[1] == this; }
  void up() { sz = c[0]->sz + c[1]->sz + 1; }
  void clear() {
    sz = 1;
    c[0] = c[1] = fa = null;
  }
};
void rot(Node* x) {
  Node* p = x->fa;
  int f = x->d();
  p->fa->setc(p->d(), x);
  p->setc(f, x->c[!f]);
  x->setc(!f, p);
  p->up();
}
void splay(Node*& root, Node* x, Node* goal) {
  while (x->fa != goal) {
    if (x->fa->fa == goal) {
      rot(x);
    } else {
      x->d() == (x->fa->d()) ? rot(x->fa) : rot(x);
      rot(x);
    }
  }
  x->up();
  if (goal == null) {
    root = x;
  }
}
void slect(Node*& root, int k, Node* goal) {
  Node* x = root;
  while (x->c[0]->sz + 1 != k) {
    if (k < x->c[0]->sz + 1) {
      x = x->c[0];
    } else {
      k -= x->c[0]->sz + 1;
      x = x->c[1];
    }
  }
  splay(root, x, goal);
}
int get_rank(Node*& root, Node* x) {
  splay(root, x, null);
  return x->c[0]->sz + 1;
}
Node pool[2 * N], *tail = pool;
Node* root;
Node* value[N];
void show(Node* root) {
  if (root == null) {
    return;
  }
  show(root->c[0]);
  printf("rank=%d\n", get_rank(::root, root->link));
  show(root->c[1]);
}
void debug(Node* x) {
  if (x->c[0] != null) debug(x->c[0]);
  printf("x = %d ", x->val);
  if (x->c[1] != null) debug(x->c[1]);
}
void erase(Node*& root, Node* x) {
  splay(root, x, null);
  Node* t = root;
  if (t->c[1] != null) {
    root = t->c[1];
    slect(root, 1, null);
    root->setc(0, t->c[0]);
  } else {
    root = root->c[0];
  }
  root->fa = null;
  if (root != null) {
    root->up();
  }
}
Node* seq_find(int pos) {
  Node* p = root;
  while (true) {
    if (p->c[0]->sz + 1 < pos) {
      pos -= p->c[0]->sz + 1;
      p = p->c[1];
    } else if (p->c[0]->sz + 1 == pos) {
      return p;
    } else {
      p = p->c[0];
    }
  }
}
void insert(Node*& root, Node* x) {
  Node* now = root;
  Node* pre = root;
  int c = -1;
  while (now != null) {
    pre = now;
    if (get_rank(::root, x->link) < get_rank(::root, now->link)) {
      now = now->c[0], c = 0;
    } else {
      now = now->c[1], c = 1;
    }
  }
  x->clear();
  pre->setc(c, x);
  splay(root, x, null);
}
void shift(int l, int r) {
  if (l == r) {
    return;
  }
  Node* x = seq_find(l);
  Node* y = seq_find(r);
  erase(root, y);
  y->clear();
  splay(root, x, null);
  if (x->c[0] == null) {
    x->setc(0, y);
    x->up();
  } else {
    Node* prev = x->c[0];
    while (prev->c[1] != null) {
      prev = prev->c[1];
    }
    splay(root, prev, x);
    prev->setc(1, y);
    prev->up();
    x->up();
  }
  erase(value[y->val], y->link);
  insert(value[y->val], y->link);
}
int get_cnt(Node*& root, int pos) {
  Node* p = root;
  Node* q = null;
  int cnt = 0;
  while (p != null) {
    q = p;
    if (pos < get_rank(::root, p->link)) {
      p = p->c[0];
    } else {
      cnt += p->c[0]->sz + 1;
      p = p->c[1];
    }
  }
  if (q != null) {
    splay(root, q, null);
  }
  return cnt;
}
void init(int n) {
  null = tail++;
  null->sz = 0;
  null->c[0] = null->c[1] = null->fa = null;
  null->val = 0;
  root = null;
  for (int i = 1; i <= n; i++) {
    value[i] = null;
  }
}
int main() {
  int n;
  static int a[N];
  scanf("%d", &n);
  init(n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 1; i <= n; i++) {
    Node* p = tail++;
    Node* q = tail++;
    p->clear();
    q->clear();
    p->setc(0, root);
    p->val = a[i];
    p->up();
    root = p;
    q->link = p, p->link = q;
    insert(value[a[i]], q);
  }
  int q, last = 0;
  scanf("%d", &q);
  while (q--) {
    int op, l, r, k;
    scanf("%d%d%d", &op, &l, &r);
    l = (l + last - 1) % n + 1;
    r = (r + last - 1) % n + 1;
    if (l > r) std::swap(l, r);
    if (op == 1) {
      shift(l, r);
    } else {
      scanf("%d", &k);
      k = (k + last - 1) % n + 1;
      last = get_cnt(value[k], r) - get_cnt(value[k], l - 1);
      printf("%d\n", last);
    }
  }
}