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>
inline int getint() {
  int x = 0, p = 1;
  char c;
  c = getchar();
  while (c <= 32) c = getchar();
  if (c == 45) p = -p, c = getchar();
  while (c > 32) x = x * 10 + c - 48, c = getchar();
  return x * p;
}
using namespace std;
const int maxn = 1e5 + 1;
const int maxs = 1000;
int n, t, ok[maxn / maxs + 1][maxn], res;
list<int> S[maxn / maxs + 1];
list<int>::iterator it;
int main() {
  n = getint();
  for (int(i) = 0; (i) < (n); (i)++) {
    int x = getint();
    S[i / maxs].push_back(x);
    ok[i / maxs][x]++;
  }
  t = getint();
  for (int(i) = 0; (i) < (t); (i)++) {
    int op, l, r, L, R;
    op = getint();
    l = getint();
    r = getint();
    L = (l + res - 1) % n + 1, R = (r + res - 1) % n + 1;
    if (L > R) swap(L, R);
    if (op == 2) {
      int k = getint(), K = (k + res - 1) % n + 1;
      L--, res = 0;
      it = S[L / maxs].begin(), l = L / maxs * maxs;
      while (l < L) l++, it++;
      while (L < R && L % maxs != 0) L++, res += (*it == K), it++;
      it = S[R / maxs].begin(), r = R / maxs * maxs;
      while (r < R) r++, it++;
      while (L < R && R % maxs != 0) R--, it--, res += (*it == K);
      while (L < R) {
        res += ok[L / maxs][K];
        L += maxs;
      }
      printf("%d\n", res);
    } else {
      L--, R--, l = L / maxs * maxs, r = R / maxs * maxs;
      it = S[R / maxs].begin();
      while (r < R) r++, it++;
      int x = *it;
      S[R / maxs].erase(it);
      ok[L / maxs][x]++;
      ok[R / maxs][x]--;
      it = S[L / maxs].begin();
      while (l < L) it++, l++;
      S[L / maxs].insert(it, x);
      l = L / maxs, r = R / maxs;
      while (l < r) {
        it = S[l].end();
        it--;
        ok[l + 1][*it]++;
        ok[l][*it]--;
        S[l + 1].push_front(*it);
        S[l].erase(it);
        l++;
      }
    }
  }
  return 0;
}