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;
const int MX = 1e5 + 9;
const int sq = 350;
int n, a[MX], block[MX];
deque<int> dq[MX];
int cnt[MX / sq + 10][MX];
void calc(int num) {
  for (int i = 0; i < dq[num].size(); i++) cnt[num][dq[num][i]]++;
}
void clear_(int num) {
  for (int i = 0; i < dq[num].size(); i++) cnt[num][dq[num][i]]--;
}
void solve(int l, int r) {
  int bl = block[l];
  if (block[l] == block[r]) {
    int last = dq[bl][l % sq];
    dq[bl][l % sq] = dq[bl][r % sq];
    for (int i = (l + 1) % sq; i <= r % sq; i++) {
      int newlast = dq[bl][i % sq];
      dq[bl][i % sq] = last;
      last = newlast;
    }
    return;
  }
  int R = block[r];
  int added = dq[R][r % sq];
  int L = block[l];
  cnt[L][added]++;
  cnt[R][added]--;
  vector<int> v;
  for (int i = 0; i < dq[block[r]].size(); i++) {
    if (i == (r % sq)) continue;
    v.push_back(dq[block[r]][i]);
  }
  dq[block[r]].clear();
  for (auto pp : v) dq[block[r]].push_back(pp);
  v.clear();
  for (int i = 0; i < (l % sq); i++) {
    v.push_back(dq[L][i]);
  }
  v.push_back(added);
  for (int i = (l % sq); i < dq[L].size(); i++) {
    v.push_back(dq[L][i]);
  }
  dq[L].clear();
  for (auto pp : v) dq[L].push_back(pp);
  v.clear();
  int last;
  int L1 = block[l];
  while (L1 < R) {
    last = dq[L1].back();
    dq[L1].pop_back();
    cnt[L1][last]--;
    cnt[L1 + 1][last]++;
    dq[L1 + 1].push_front(last);
    L1++;
  }
}
int getcntnum(int l, int r, int x) {
  int bl = block[l];
  int ans = 0;
  if (block[l] == block[r]) {
    for (int i = l; i <= r; i++) {
      ans += (dq[block[l]][i % sq] == x);
    }
    return ans;
  }
  while (block[l] == bl && l < n) {
    ans += (dq[bl][l % sq] == x);
    l++;
  }
  bl = block[r];
  while (block[r] == bl && r) {
    ans += (dq[bl][r % sq] == x);
    r--;
  }
  int L = block[l], R = block[r];
  while (L <= R) {
    ans += cnt[L][x];
    L++;
  }
  return ans;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 0; i < n; i++) {
    int block1 = i / sq;
    dq[block1].push_back(a[i]);
    cnt[block1][a[i]]++;
    block[i] = i / sq;
  }
  int ans = 0;
  int q;
  cin >> q;
  while (q--) {
    int t;
    scanf("%d", &t);
    if (t == 1) {
      int l, r;
      scanf("%d%d", &l, &r);
      l = (l + ans - 1) % n;
      r = (r + ans - 1) % n;
      if (l > r) swap(l, r);
      solve(l, r);
    } else {
      int l, r, x;
      scanf("%d%d%d", &l, &r, &x);
      l = (l + ans - 1) % n;
      r = (r + ans - 1) % n;
      if (l > r) swap(l, r);
      x = (x + ans - 1) % n;
      x++;
      ans = getcntnum(l, r, x);
      printf("%d\n", ans);
    }
  }
}