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;
int n, m, ans, t;
int L[510], R[510], dr[510], dt[510], bel[100010], lst[200010], nxt[200010],
    cnt[320][100010], a[100010];
int read() {
  int r = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c <= '9' && c >= '0') r = (r << 1) + (r << 3) + c - '0', c = getchar();
  return r * f;
}
void put(int x) {
  if (x < 0) {
    putchar('-');
    x = ~x + 1;
  }
  if (x > 9) put(x / 10);
  putchar(x % 10 + '0');
}
void make_blocks() {
  t = sqrt(n);
  for (int i = 1; i <= n; i++) {
    bel[i] = (i + t - 1) / t;
    if (!L[bel[i]]) L[bel[i]] = i;
    R[bel[i]] = i;
  }
  for (int i = 1; i <= bel[n]; i++) {
    dr[i] = n + i * 2 - 1, dt[i] = n + i * 2;
    for (int j = L[i]; j <= R[i]; j++)
      cnt[i][a[j]]++, nxt[j] = j + 1, lst[j] = j - 1;
    lst[L[i]] = dr[i], nxt[dr[i]] = L[i], nxt[R[i]] = dt[i], lst[dt[i]] = R[i];
  }
}
void change(int l, int r) {
  if (l == r) return;
  if (bel[l] == bel[r]) {
    int e = bel[l], x = dr[e], y = dt[e];
    for (int i = L[e]; i <= l; i++) x = nxt[x];
    for (int i = R[e]; i >= r; i--) y = lst[y];
    lst[nxt[y]] = lst[y], nxt[lst[y]] = nxt[y];
    nxt[lst[x]] = y, lst[y] = lst[x];
    lst[x] = y, nxt[y] = x;
    return;
  }
  int el = bel[l], er = bel[r], x = dt[el], y = dr[er];
  for (int i = R[el]; i >= l; i--) x = lst[x];
  for (int i = L[er]; i <= r; i++) y = nxt[y];
  cnt[el][a[y]]++, cnt[er][a[y]]--;
  lst[nxt[y]] = lst[y], nxt[lst[y]] = nxt[y];
  nxt[lst[x]] = y, lst[y] = lst[x];
  lst[x] = y, nxt[y] = x;
  for (int i = el; i < er; i++) {
    int x = nxt[dr[i + 1]], y = lst[dt[i]];
    cnt[i][a[y]]--, cnt[i + 1][a[y]]++;
    lst[nxt[y]] = lst[y], nxt[lst[y]] = nxt[y];
    nxt[lst[x]] = y, lst[y] = lst[x];
    lst[x] = y, nxt[y] = x;
  }
}
int query(int l, int r, int k) {
  if (bel[l] == bel[r]) {
    int e = bel[l], x = dr[e], y = dt[e], as = 0;
    for (int i = L[e]; i <= l; i++) x = nxt[x];
    for (int i = R[e]; i >= r; i--) y = lst[y];
    while (x ^ y) {
      if (a[x] == k) as++;
      x = nxt[x];
    }
    if (a[y] == k) as++;
    return as;
  }
  int el = bel[l], er = bel[r], x = dt[el], y = dr[er], sum = 0;
  for (int i = R[el]; i >= l; i--) {
    x = lst[x];
    if (a[x] == k) sum++;
  }
  for (int i = L[er]; i <= r; i++) {
    y = nxt[y];
    if (a[y] == k) sum++;
  }
  for (int i = el + 1; i < er; i++) sum += cnt[i][k];
  return sum;
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  make_blocks();
  m = read();
  while (m--) {
    int opt = read(), ll = (read() + ans - 1) % n + 1,
        rr = (read() + ans - 1) % n + 1;
    if (ll > rr) swap(ll, rr);
    if (opt == 1)
      change(ll, rr);
    else {
      int kk = (read() + ans - 1) % n + 1;
      ans = query(ll, rr, kk);
      put(ans);
      puts("");
    }
  }
  return 0;
}