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;
int a[100010];
int sq = 300;
int sq1;
deque<int> dq[400];
int rep[350][100010];
void spin(int l, int r) {
  int z = l / sq;
  int l1 = l % sq;
  int r1 = r % sq;
  int qwe = dq[z][r1];
  for (int i = r1; i > l1; i--) {
    dq[z][i] = dq[z][i - 1];
  }
  dq[z][l1] = qwe;
}
void spinl(int l, int r) {
  int z = l / sq;
  int qwe = dq[r / sq][r % sq];
  int l1 = l % sq;
  int r1 = dq[z].size() - 1;
  rep[z][dq[z][r1]]--;
  rep[z][qwe]++;
  for (int i = r1; i > l1; i--) {
    dq[z][i] = dq[z][i - 1];
  }
  dq[z][l1] = qwe;
}
void spinr(int l, int r, int lr) {
  int z = r / sq;
  int l1 = 0;
  int r1 = r % sq;
  rep[z][dq[z][r1]]--;
  rep[z][lr]++;
  for (int i = r1; i > l1; i--) {
    dq[z][i] = dq[z][i - 1];
  }
  dq[z][l1] = lr;
}
int calc(int l, int r, int k) {
  int z = l / sq;
  int l1 = l % sq;
  int r1 = r % sq;
  int res = 0;
  for (int i = r1; i >= l1; i--) {
    if (dq[z][i] == k) res++;
  }
  return res;
}
int calcl(int l, int r, int k) {
  int z = l / sq;
  int l1 = l % sq;
  int r1 = dq[z].size() - 1;
  int res = 0;
  for (int i = r1; i >= l1; i--) {
    if (dq[z][i] == k) res++;
  }
  return res;
}
int calcr(int l, int r, int k) {
  int z = r / sq;
  int l1 = 0;
  int r1 = r % sq;
  int res = 0;
  for (int i = r1; i >= l1; i--) {
    if (dq[z][i] == k) res++;
  }
  return res;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    dq[i / sq].push_back(a[i]);
    rep[i / sq][a[i]]++;
  }
  sq1 = (n - 1) / sq + 1;
  int q;
  cin >> q;
  int lastans = 0;
  while (q--) {
    int l, r, k, t;
    cin >> t;
    if (t == 1) {
      cin >> l >> r;
      l = ((l + lastans - 1) % n);
      r = ((r + lastans - 1) % n);
      if (l > r) swap(l, r);
      int l1 = l / sq;
      int r1 = r / sq;
      if (l1 == r1) {
        spin(l, r);
        continue;
      }
      int lr = dq[l1].back();
      spinl(l, r);
      for (int i = l1 + 1; i < r1; i++) {
        rep[i][lr]++;
        dq[i].push_front(lr);
        lr = dq[i].back();
        rep[i][lr]--;
        dq[i].pop_back();
      }
      spinr(l, r, lr);
    } else {
      cin >> l >> r >> k;
      l = ((l + lastans - 1) % n);
      r = ((r + lastans - 1) % n);
      k = ((k + lastans - 1) % n) + 1;
      if (l > r) swap(l, r);
      int l1 = l / sq;
      int r1 = r / sq;
      long long ans = 0;
      if (l1 == r1) {
        ans = calc(l, r, k);
        lastans = ans;
        cout << ans << endl;
        continue;
      }
      ans += calcl(l, r, k);
      for (int i = l1 + 1; i < r1; i++) {
        ans += rep[i][k];
      }
      ans += calcr(l, r, k);
      lastans = ans;
      cout << ans << endl;
    }
  }
  return 0;
}