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;
template <class A, class B>
ostream& operator<<(ostream& out, const pair<A, B>& a) {
  return out << "(" << a.first << ", " << a.second << ")";
}
template <class A>
ostream& operator<<(ostream& out, const vector<A>& a) {
  out << "[";
  for (auto it = a.begin(); it != a.end(); ++it) {
    if (it != a.begin()) out << ", ";
    out << *it;
  }
  return out << "]";
}
const int INF = 1e9;
const long long INF64 = 1e18;
const int MOD = 1e9 + 7;
const long double PI = acosl(-1.0);
const long double EPS = 1e-9;
mt19937 rnd(time(NULL));
mt19937_64 rnd64(time(NULL));
const int N = 100 * 1000 + 11;
const int SQN = 330;
const int NSQN = 330;
int n;
vector<int> a;
int cnt[NSQN][N];
vector<int> s[NSQN];
bool read() {
  if (scanf("%d", &n) != 1) return false;
  a.resize(n);
  for (int i = 0; i < int(n); ++i) scanf("%d", &a[i]);
  return true;
}
void rebuild() {
  vector<int> na;
  for (int i = 0; i < int(NSQN); ++i) {
    for (auto it : s[i]) {
      na.push_back(it);
      --cnt[i][it];
    }
    s[i].clear();
  }
  for (int i = 0; i < int(n); ++i) {
    int cur = i / SQN;
    s[cur].push_back(na[i]);
    ++cnt[cur][na[i]];
  }
}
int get(int l, int r, int k) {
  int res = 0;
  int cursz = 0;
  for (int i = 0; i < int(NSQN); ++i) {
    if (s[i].empty()) continue;
    if (cursz < l) {
      if (cursz + int((s[i]).size()) >= l) {
        for (auto it : s[i]) {
          ++cursz;
          if (cursz >= l && cursz <= r) res += (it == k);
        }
      } else {
        cursz += int((s[i]).size());
      }
    } else {
      if (cursz > r) break;
      if (cursz + int((s[i]).size()) <= r) {
        res += cnt[i][k];
        cursz += int((s[i]).size());
      } else {
        for (auto it : s[i]) {
          ++cursz;
          if (cursz <= r)
            res += (it == k);
          else
            break;
        }
      }
    }
  }
  return res;
}
void upd(int l, int r) {
  int cursz = 0;
  int val = -1;
  bool fl = false;
  for (int i = 0; i < int(NSQN); ++i) {
    if (fl) break;
    if (cursz + int((s[i]).size()) >= r) {
      for (int j = 0; j < int(int((s[i]).size())); ++j) {
        cursz++;
        if (cursz == r) {
          val = s[i][j];
          --cnt[i][val];
          s[i].erase(s[i].begin() + j);
          fl = true;
          break;
        }
      }
    } else {
      cursz += int((s[i]).size());
    }
  }
  assert(val != -1);
  fl = false;
  cursz = 0;
  for (int i = 0; i < int(NSQN); ++i) {
    if (fl) break;
    if (cursz + int((s[i]).size()) >= l) {
      for (int j = 0; j < int(int((s[i]).size())); ++j) {
        cursz++;
        if (cursz == l) {
          ++cnt[i][val];
          s[i].insert(s[i].begin() + j, val);
          fl = true;
          return;
        }
      }
    } else {
      cursz += int((s[i]).size());
    }
  }
  s[NSQN - 1].push_back(val);
  ++cnt[NSQN - 1][val];
}
void print() {
  for (int i = 0; i < int(NSQN); ++i)
    for (auto it : s[i]) cerr << it << " ";
  cerr << endl;
}
void solve() {
  for (int i = 0; i < int(n); ++i) {
    int cur = i / SQN;
    s[cur].push_back(a[i]);
    ++cnt[cur][a[i]];
  }
  int q;
  scanf("%d", &q);
  int lastans = 0;
  for (int i = 0; i < int(q); ++i) {
    if (i % SQN == 0) rebuild();
    int tp, ll, rr;
    scanf("%d %d %d", &tp, &ll, &rr);
    int l = (ll + lastans - 1) % n + 1;
    int r = (rr + lastans - 1) % n + 1;
    if (l > r) swap(l, r);
    if (tp == 1) {
      upd(l, r);
    } else {
      int kk;
      scanf("%d", &kk);
      int k = (kk + lastans - 1) % n + 1;
      lastans = get(l, r, k);
      printf("%d\n", lastans);
    }
  }
}
int main() {
  cout << fixed << setprecision(10);
  cerr << fixed << setprecision(10);
  while (read()) {
    solve();
  }
}