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 G = 300;
const int GC = N / G + 10;
const int P = 507;
int n, m, c, k, t, l, p, lit, x, y, z, ans;
vector<int> a, v[GC];
void C(int& x) { x = (x + ans - 1) % n + 1; }
int s[GC][N];
void A(int i, int x, int t) {
  v[i].insert(v[i].begin() + x, t);
  s[i][t]++;
}
void D(int i, int x) {
  t = v[i][x - 1];
  v[i].erase(v[i].begin() + (x - 1));
  s[i][t]--;
}
void R() {
  for (int i = 1; i <= c; i++) {
    for (int k = 0; k < v[i].size(); k++)
      s[i][v[i][k]] = 0, a.push_back(v[i][k]);
    v[i].clear();
  }
  for (int id = 1, i = 0; i < n; i += G, id++)
    for (int j = i; j < i + G && j < n; j++) A(id, v[id].size(), a[j]);
  a.clear();
}
int S(int x, int k) {
  int i, j, p, ans = 0;
  for (p = 0, i = 1; p + v[i].size() < x; p += v[i].size(), i++) ans += s[i][k];
  for (j = 0; j < x - p; j++) ans += (v[i][j] == k);
  return ans;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &x), a.push_back(x);
  c = (n / G) + (n % G > 0);
  R();
  for (scanf("%d", &m); m; m--) {
    scanf("%d", &z);
    if (z == 1) {
      scanf("%d%d", &x, &y);
      C(x);
      C(y);
      if (x > y) swap(x, y);
      if (x == y) continue;
      int i, p;
      if (++lit >= G) lit = 0, R();
      for (p = 0, i = 1; p + v[i].size() < y; p += v[i].size(), i++)
        ;
      D(i, y - p);
      for (p = 0, i = 1; p + v[i].size() < x; p += v[i].size(), i++)
        ;
      A(i, x - p - 1, t);
    } else {
      scanf("%d%d%d", &x, &y, &k);
      C(x);
      C(y);
      C(k);
      if (x > y) swap(x, y);
      printf("%d\n", ans = S(y, k) - S(x - 1, k));
    }
  }
  return 0;
}