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
#include <bits/stdc++.h>
using namespace std;
template <typename A, typename B>
inline void chmin(A &a, B b) {
  if (a > b) a = b;
}
template <typename A, typename B>
inline void chmax(A &a, B b) {
  if (a < b) a = b;
}
const long long MN = 100000;
const long long B = 350;
long long N, Q;
long long A[MN];
deque<long long> d[(MN + B - 1) / B];
long long cnt[(MN + B - 1) / B][100000];
signed main() {
  scanf("%lld", &N);
  for (long long i = 0; i < (N); i++) {
    scanf("%lld", &A[i]);
    A[i]--;
    d[i / B].push_back(A[i]);
    cnt[i / B][A[i]]++;
  }
  scanf("%lld", &Q);
  long long lastans = 0;
  while (Q--) {
    long long t;
    long long l, r, k;
    scanf("%lld%lld%lld", &t, &l, &r);
    if (t == 2) scanf("%lld", &k);
    l = (l + lastans - 1 + N) % N + 1;
    r = (r + lastans - 1 + N) % N + 1;
    k = (k + lastans - 1 + N) % N + 1;
    if (l > r) swap(l, r);
    l--;
    k--;
    if (t == 1) {
      if (l / B == r / B) {
        long long ll = l % B;
        long long rr = r % B;
        rotate(d[l / B].begin() + ll, d[l / B].begin() + rr - 1,
               d[l / B].begin() + rr);
        continue;
      }
      long long pre = d[l / B].back();
      d[l / B].pop_back();
      cnt[l / B][pre]--;
      for (long long i = l / B + 1; i * B < r; i++) {
        if ((i + 1) * B <= r) {
          cnt[i][pre]++;
          d[i].push_front(pre);
          pre = d[i].back();
          d[i].pop_back();
          cnt[i][pre]--;
        } else {
          cnt[i][pre]++;
          d[i].push_front(pre);
          pre = d[i][r % B];
          d[i].erase(d[i].begin() + r % B);
          cnt[i][pre]--;
        }
      }
      cnt[l / B][pre]++;
      d[l / B].insert(d[l / B].begin() + l % B, pre);
    } else {
      lastans = 0;
      for (long long i = l / B; i * B < r; i++) {
        if (l <= i * B && (i + 1) * B <= r)
          lastans += cnt[i][k];
        else {
          for (long long j = 0; j < d[i].size(); j++) {
            if (l <= i * B + j && i * B + j < r) lastans += (d[i][j] == k);
          }
        }
      }
      printf("%lld\n", lastans);
    }
  }
}
