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 mxn = 1e5 + 5;
const int sqn = 320;
int n, q;
int a[mxn];
bool flag = 0;
struct bucket {
  int cnt[mxn], sz, elem[sqn * 2 + 5];
  void clean() {
    while (sz) {
      cnt[elem[--sz]]--;
    }
    return;
  }
  void add(int pos, int x) {
    cnt[x]++;
    pos--;
    for (int i = sz - 1; i >= pos; i--) {
      elem[i + 1] = elem[i];
    }
    sz++;
    elem[pos] = x;
  }
  int del(int pos) {
    int ret = elem[--pos];
    for (int i = pos; i < sz - 1; i++) {
      elem[i] = elem[i + 1];
    }
    sz--;
    cnt[ret]--;
    return ret;
  }
  int query(int x, int y, int k) {
    if (x == 1 && y == sz) return cnt[k];
    int ret = 0;
    for (int i = x - 1; i < y; i++) ret += elem[i] == k;
    return ret;
  }
} bkt[sqn];
void init() {
  for (int i = 0; i < sqn; i++) {
    bkt[i].clean();
  }
  for (int i = 0; i < n; i++) {
    bkt[i / sqn].add(i % sqn + 1, a[i]);
  }
}
void reset() {
  int ptr = 0;
  for (int i = 0; i < sqn; i++)
    for (int j = 0; j < bkt[i].sz; j++) a[ptr++] = bkt[i].elem[j];
  flag = 0;
}
void add(int pos, int x) {
  for (int i = 0; i < sqn; i++) {
    if (pos <= bkt[i].sz) {
      bkt[i].add(pos, x);
      if (bkt[i].sz >= 2 * sqn) flag = 1;
      return;
    } else {
      pos -= bkt[i].sz;
    }
  }
}
int del(int pos) {
  for (int i = 0; i < sqn; i++) {
    if (pos <= bkt[i].sz) {
      return bkt[i].del(pos);
    } else {
      pos -= bkt[i].sz;
    }
  }
  return 0;
}
int query(int x, int y, int k) {
  int u = 0, ret = 0;
  for (int i = 0; i < sqn; i++) {
    int st = u + 1;
    int en = u + bkt[i].sz;
    st = max(st, x);
    en = min(en, y);
    if (st <= en) ret += bkt[i].query(st - u, en - u, k);
    u += bkt[i].sz;
  }
  return ret;
}
signed main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  init();
  scanf("%d", &q);
  int lastans = 0;
  while (q--) {
    int t, le, ri, k;
    scanf("%d", &t);
    if (t == 1) {
      scanf("%d%d", &le, &ri);
    } else {
      scanf("%d%d%d", &le, &ri, &k);
    }
    le = (le + lastans - 1 + n) % n + 1;
    ri = (ri + lastans - 1 + n) % n + 1;
    k = (k + lastans - 1 + n) % n + 1;
    if (le > ri) swap(le, ri);
    if (t == 1) {
      int deleted = del(ri);
      add(le, deleted);
    } else {
      lastans = query(le, ri, k);
      printf("%d\n", lastans);
    }
    if (flag) {
      reset();
      init();
    }
  }
}