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 BS;
int cnt[400][100005];
deque<int> dq[400];
int n;
void print() {
  for (int i = 0; i <= (n - 1) % BS; ++i) {
    for (int j = 0; j < dq[i].size(); ++j) printf("%d ", dq[i][j]);
    printf("\n");
  }
}
int getAns(int l, int r, int k) {
  int ans = 0;
  int stB = l / BS + 1, enB = r / BS - 1;
  for (int i = stB; i <= enB; ++i) ans += cnt[i][k];
  int f = l % BS;
  int en = r % BS;
  if (l / BS == r / BS) {
    int idx = l / BS;
    for (int i = f; i <= en; ++i) ans += dq[idx][i] == k;
  } else {
    int idx = l / BS;
    for (int i = f; i < dq[idx].size(); ++i) ans += dq[idx][i] == k;
    idx = r / BS;
    for (int i = 0; i <= en; ++i) ans += dq[idx][i] == k;
  }
  return ans;
}
void update(int l, int r) {
  int lB = l / BS, st = l % BS;
  int rB = r / BS, en = r % BS;
  int x = dq[rB][en];
  dq[rB].erase(dq[rB].begin() + en);
  dq[lB].insert(dq[lB].begin() + st, x);
  cnt[lB][x]++;
  cnt[rB][x]--;
  for (int i = lB; i < rB; ++i) {
    int siz = dq[i].size();
    x = dq[i][siz - 1];
    dq[i + 1].push_front(x);
    dq[i].pop_back();
    cnt[i][x]--;
    cnt[i + 1][x]++;
  }
}
int main() {
  cin >> n;
  BS = max(1, (int)sqrt(n));
  for (int i = 0; i < n; ++i) {
    int x;
    scanf("%d", &x);
    int BI = i / BS;
    cnt[BI][x]++;
    dq[BI].push_back(x);
  }
  int q;
  cin >> q;
  int lastAns = 0;
  while (q--) {
    int t, l, r;
    scanf("%d%d%d", &t, &l, &r);
    l = (l + lastAns - 1) % n + 1;
    r = (r + lastAns - 1) % n + 1;
    l--, r--;
    if (l > r) swap(l, r);
    if (t == 1) {
      update(l, r);
    } else {
      int k;
      scanf("%d", &k);
      k = (k + lastAns - 1) % n + 1;
      lastAns = getAns(l, r, k);
      printf("%d\n", lastAns);
    }
  }
  return 0;
}