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, Q, T, L, R, K, lastans;
int A[100000];
const int B_N = 250;
const int B_SZ = 400;
int sz[B_N];
vector<int> lis[B_N];
vector<int> st[B_N];
void init() {
  memset((sz), (0), sizeof(sz));
  int ptr = 0;
  for (int(i) = 0; (i) < (B_N); (i)++)
    for (int(j) = 0; (j) < (lis[i].size()); (j)++) A[ptr++] = lis[i][j];
  for (int(i) = 0; (i) < (B_N); (i)++) lis[i].clear(), st[i].clear();
  for (int(i) = 0; (i) < (N); (i)++)
    lis[i / B_SZ].push_back(A[i]), sz[i / B_SZ]++, st[i / B_SZ].push_back(A[i]);
  for (int(i) = 0; (i) < (B_N); (i)++) sort((st[i]).begin(), (st[i]).end());
}
int st_erase(int pos) {
  int zan = 0;
  for (int(i) = 0; (i) < (B_N); (i)++) {
    if (zan <= pos && pos < zan + sz[i]) {
      int ss = pos - zan;
      int tt = lis[i][ss];
      lis[i].erase(lis[i].begin() + ss);
      st[i].erase(lower_bound((st[i]).begin(), (st[i]).end(), tt));
      sz[i]--;
      return tt;
    }
    zan += sz[i];
  }
}
void st_insert(int pos, int val) {
  int zan = 0;
  for (int(i) = 0; (i) < (B_N); (i)++) {
    if (zan <= pos && pos < zan + sz[i]) {
      int ss = pos - zan;
      lis[i].insert(lis[i].begin() + ss, val);
      st[i].insert(lower_bound((st[i]).begin(), (st[i]).end(), val), val);
      sz[i]++;
      return;
    }
    zan += sz[i];
  }
}
int solve(int l, int r, int k) {
  int res = 0, zan = 0;
  for (int(i) = 0; (i) < (B_N); (i)++) {
    if (l <= zan && zan + sz[i] <= r) {
      res += upper_bound((st[i]).begin(), (st[i]).end(), (k)) -
             lower_bound((st[i]).begin(), (st[i]).end(), (k));
    } else if (r <= zan || zan + sz[i] <= l) {
      zan += sz[i];
      continue;
    } else {
      for (int(j) = 0; (j) < (sz[i]); (j)++) {
        int pos = zan + j;
        if (l <= pos && pos < r && lis[i][j] == k) res++;
      }
    }
    zan += sz[i];
  }
  return res;
}
int main() {
  scanf("%d", &N);
  for (int(i) = 0; (i) < (N); (i)++) scanf("%d", &A[i]);
  init();
  scanf("%d", &Q);
  for (int(i) = 0; (i) < (Q); (i)++) {
    if (i % 300 == 299) init();
    scanf("%d", &T);
    if (T == 1) {
      scanf("%d%d", &L, &R);
      L = (L + lastans - 1) % N + 1;
      R = (R + lastans - 1) % N + 1;
      if (L > R) swap(L, R);
      L--;
      int tmp = st_erase(R - 1);
      st_insert(L, tmp);
    } else {
      scanf("%d%d%d", &L, &R, &K);
      L = (L + lastans - 1) % N + 1;
      R = (R + lastans - 1) % N + 1;
      K = (K + lastans - 1) % N + 1;
      if (L > R) swap(L, R);
      L--;
      int res = solve(L, R, K);
      printf("%d\n", res);
      lastans = res;
    }
  }
}