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 long double EPS = (long double)1e-7;
const long double PI = acos(0) * 2;
bool isZero(const long double& x) { return abs(x) <= EPS; }
int sign(const long double& x) { return isZero(x) ? 0 : (0 < x ? 1 : -1); }
long long gcd(long long a, long long b) {
  for (; b; a %= b, swap(a, b)) {
  }
  return abs(a);
}
pair<long long, long long> operator+(const pair<long long, long long>& a,
                                     const pair<long long, long long>& b) {
  return pair<long long, long long>(a.first + b.first, a.second + b.second);
}
pair<long long, long long> operator-(const pair<long long, long long>& a,
                                     const pair<long long, long long>& b) {
  return pair<long long, long long>(a.first - b.first, a.second - b.second);
}
pair<long long, long long> operator*(const pair<long long, long long>& a,
                                     const long long& b) {
  return pair<long long, long long>(a.first * b, a.second * b);
}
long long operator*(const pair<long long, long long>& a,
                    const pair<long long, long long>& b) {
  return a.first * b.second - b.first * a.second;
}
long long ccw(const pair<long long, long long>& a,
              const pair<long long, long long>& b,
              const pair<long long, long long>& c) {
  return a * b + b * c + c * a;
}
void fg(vector<int> G[], int a, int b) {
  G[a].emplace_back(b);
  G[b].emplace_back(a);
}
void fg(vector<pair<int, int> > G[], int a, int b, int c) {
  G[a].emplace_back(b, c);
  G[b].emplace_back(a, c);
}
const int MAXN = 100005;
const int SQRN = 320;
int A[SQRN][MAXN];
deque<int> B[SQRN];
int N, Q;
void f(int p, int q) {
  if (p >= q) return;
  int pi = p / SQRN, qi = q / SQRN;
  int pj = p % SQRN, qj = q % SQRN;
  if (pi == qi) {
    rotate(B[pi].begin() + pj, B[pi].begin() + qj, B[pi].begin() + qj + 1);
    return;
  }
  for (int i = pi, x; i < qi; i++) {
    x = B[i].back();
    B[i].pop_back();
    B[i + 1].emplace_front(x);
    A[i][x]--;
    A[i + 1][x]++;
  }
  {
    int x = B[qi][qj + 1];
    B[qi].erase(B[qi].begin() + qj + 1);
    B[pi].insert(B[pi].begin() + pj, x);
    A[qi][x]--;
    A[pi][x]++;
  }
}
int f(int p, int q, int x) {
  int pi = p / SQRN, qi = q / SQRN;
  int pj = p % SQRN, qj = q % SQRN;
  int ret = 0;
  if (pi == qi) {
    for (int i = pj; i <= qj; i++)
      if (B[pi][i] == x) ret++;
    return ret;
  }
  for (int i = ((int)(B[pi]).size()) - 1; pj <= i; i--)
    if (B[pi][i] == x) ret++;
  for (int i = qj; 0 <= i; i--)
    if (B[qi][i] == x) ret++;
  for (int i = pi + 1; i < qi; i++) ret += A[i][x];
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> N;
  for (int i = 0, c, bi; i < N; i++) {
    cin >> c;
    bi = i / SQRN;
    A[bi][c]++;
    B[bi].emplace_back(c);
  }
  cin >> Q;
  for (int t, a, b, c, pv = 0; Q--;) {
    cin >> t >> a >> b;
    if (2 == t) cin >> c;
    a = (a + pv - 1) % N;
    b = (b + pv - 1) % N;
    c = (c + pv - 1) % N + 1;
    if (a > b) swap(a, b);
    if (1 == t)
      f(a, b);
    else
      cout << (pv = f(a, b, c)) << '\n';
  }
  return 0;
}