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 inf = 1e9 + 7;
template <class T>
inline T read() {
  char ch;
  bool flag = 0;
  T x = 0;
  while (ch = getchar(), !isdigit(ch))
    if (ch == '-') flag = 1;
  while (isdigit(ch)) x = x * 10 + ch - 48, ch = getchar();
  return flag ? -x : x;
}
struct {
  inline operator int() { return read<int>(); }
  template <class T>
  inline void operator()(T &x) {
    x = *this;
  }
  template <class T, class... A>
  inline void operator()(T &x, A &...a) {
    x = *this, this->operator()(a...);
  }
} IN;
template <class T>
inline void chkmax(T &x, T y) {
  if (x < y) x = y;
}
template <class T>
inline void chkmin(T &x, T y) {
  if (x > y) x = y;
}
const int N = 1e5 + 5;
int n, m, a[N];
const int S = 327;
int sqrtn, L[S], R[S], t[S], id[N], val[S][N], sta[S][N + S];
inline void init() {
  sqrtn = sqrt(n + 1);
  for (int l = 1, c = 1, r; l <= n; l = r + 1, ++c) {
    L[c] = l, r = R[c] = min(l + sqrtn, n), t[c] = N;
    for (int i = l; i <= r; ++i) id[i] = c;
    for (int i = l; i <= r; ++i)
      sta[c][t[c] + i - L[c] + 1] = a[i], ++val[c][a[i]];
  }
}
inline void modify(int l, int r) {
  if (id[l] == id[r]) {
    int id = ::id[l], tmp = sta[id][t[id] + r - L[id] + 1];
    for (int i = r - 1; i >= l; --i)
      sta[id][t[id] + i + 1 - L[id] + 1] = sta[id][t[id] + i - L[id] + 1];
    return sta[id][t[id] + l - L[id] + 1] = tmp, void();
  }
  int idl = id[l], idr = id[r], now = sta[idl][t[idl] + R[idl] - L[idl] + 1],
      tmp = sta[idr][t[idr] + r - L[idr] + 1];
  --val[idl][now], ++val[idl][tmp], --val[idr][tmp];
  for (int i = R[idl] - 1; i >= l; --i)
    sta[idl][t[idl] + i + 1 - L[idl] + 1] = sta[idl][t[idl] + i - L[idl] + 1];
  for (int i = id[l] + 1; i <= id[r] - 1; ++i)
    sta[i][t[i]] = now, ++val[i][now], now = sta[i][t[i] + R[i] - L[i] + 1],
    --val[i][now], --t[i];
  for (int i = r - 1; i >= L[idr]; --i)
    sta[idr][t[idr] + i + 1 - L[idr] + 1] = sta[idr][t[idr] + i - L[idr] + 1];
  sta[idr][t[idr] + L[idr] - L[idr] + 1] = now,
                                      sta[idl][t[idl] + l - L[idl] + 1] = tmp,
                                      ++val[idr][now];
}
inline int query(int l, int r, int k) {
  int res = 0;
  if (id[l] == id[r]) {
    for (int i = l; i <= r; ++i)
      res += (sta[id[l]][t[id[l]] + i - L[id[l]] + 1] == k);
    return res;
  }
  for (int i = l; i <= R[id[l]]; ++i)
    res += (sta[id[l]][t[id[l]] + i - L[id[l]] + 1] == k);
  for (int i = L[id[r]]; i <= r; ++i)
    res += (sta[id[r]][t[id[r]] + i - L[id[r]] + 1] == k);
  for (int i = id[l] + 1; i <= id[r] - 1; ++i) res += val[i][k];
  return res;
}
int l, r, op, lans;
int main() {
  IN(n);
  for (int i = 1; i <= n; ++i) IN(a[i]);
  init();
  IN(m);
  while (m--) {
    IN(op, l, r), l = (l + lans - 1) % n + 1, r = (r + lans - 1) % n + 1;
    if (l > r) swap(l, r);
    if (op == 1) modify(l, r);
    if (op == 2)
      printf("%d\n", lans = query(l, r, (int(IN) + lans - 1) % n + 1));
  }
  return 0;
}