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 = 100010;
const int S = 550;
int n, q, a[N];
struct Block {
  int lid, rid;
  deque<int> que;
  int mp[N];
  Block() {}
  void init() {
    lid = rid = 0;
    que.clear();
    memset(mp, 0, sizeof mp);
  }
  void addf(int x) {
    que.push_front(x);
    mp[x]++;
  }
  void addb(int x) {
    que.push_back(x);
    mp[x]++;
  }
  void delf() {
    if (que.empty()) return;
    int x = que.front();
    que.pop_front();
    mp[x]--;
  }
  void delb() {
    if (que.empty()) return;
    int x = que.back();
    que.pop_back();
    mp[x]--;
  }
} all[S];
int tot;
void init() {
  for (int i = 0; i < S; ++i) all[i].init();
  int id = 1;
  tot = 0;
  for (int i = 0; id <= n; ++i) {
    tot++;
    all[i].lid = all[i].rid = id;
    for (int j = 1; j <= S && id <= n; ++j) {
      all[i].addb(a[id]);
      all[i].rid++;
      id++;
    }
  }
  for (int i = 0; i < tot; ++i) all[i].rid--;
}
int st[N], top;
void mov(int l, int r) {
  int sx = 0, tx = 0;
  for (int i = 0; i < tot; ++i) {
    if (all[i].lid <= l && l <= all[i].rid) {
      sx = i;
    }
    if (all[i].lid <= r && r <= all[i].rid) {
      tx = i;
    }
  }
  for (int i = sx + 1; i <= tx; ++i) {
    all[i].addf(all[i - 1].que.back());
    all[i - 1].delb();
  }
  int val = 0;
  top = 0;
  for (int i = all[tx].rid; i > r; --i) {
    st[top++] = all[tx].que.back();
    all[tx].delb();
  }
  val = all[tx].que.back();
  all[tx].delb();
  while (top > 0) all[tx].addb(st[--top]);
  for (int i = all[sx].lid; i < l; ++i) {
    st[top++] = all[sx].que.front();
    all[sx].delf();
  }
  all[sx].addf(val);
  while (top > 0) all[sx].addf(st[--top]);
}
int cal(int ip, int l, int r, int k) {
  int res = 0;
  top = 0;
  for (int i = 1; i <= l - 1; ++i) {
    st[top++] = all[ip].que.front();
    all[ip].delf();
  }
  int alx = all[ip].rid - all[ip].lid + 1;
  for (int i = 1; i <= alx - r; ++i) {
    st[top++] = all[ip].que.back();
    all[ip].delb();
  }
  res = all[ip].mp[k];
  for (int i = 1; i <= alx - r; ++i) {
    all[ip].addb(st[--top]);
  }
  for (int i = 1; i <= l - 1; ++i) {
    all[ip].addf(st[--top]);
  }
  return res;
}
int qry(int l, int r, int k) {
  int res = 0;
  for (int i = 0; i < tot; ++i) {
    if (all[i].lid <= l && r <= all[i].rid)
      return cal(i, l - all[i].lid + 1, r - all[i].lid + 1, k);
    if (l <= all[i].lid && all[i].rid <= r) res += all[i].mp[k];
    if (all[i].lid < l && l <= all[i].rid) {
      res += cal(i, l - all[i].lid + 1, all[i].rid - all[i].lid + 1, k);
    }
    if (all[i].lid <= r && all[i].rid > r) {
      res += cal(i, 1, r - all[i].lid + 1, k);
    }
  }
  return res;
}
int scan() {
  int res = 0;
  char ch;
  while ((ch = getchar()) && (ch != ' ' && ch != '\n'))
    res = res * 10 + ch - '0';
  return res;
}
int Run() {
  n = scan();
  for (int i = 1; i <= n; ++i) a[i] = scan();
  init();
  int lans = 0;
  q = scan();
  while (q--) {
    int op, l, r;
    op = scan(), l = scan(), r = scan();
    l = (l + lans - 1) % n + 1;
    r = (r + lans - 1) % n + 1;
    if (l > r) swap(l, r);
    if (op == 1) {
      mov(l, r);
    } else {
      int k;
      k = scan();
      k = (k + lans - 1) % n + 1;
      lans = qry(l, r, k);
      printf("%d\n", lans);
    }
  }
  return 0;
}
int main() { return Run(); }