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 B = 400;
const int W = 260;
const int NMAX = B * W + 1;
int A[NMAX];
int C[W][NMAX];
vector<int> V[W];
void re_build(int N) {
  int c = 0;
  for (int i = (int)(0); i < (int)(W); i++)
    for (int j = (int)(0); j < (int)(V[i].size()); j++) A[c++] = V[i][j];
  assert(c == N);
  for (int i = (int)(0); i < (int)(W); i++) {
    for (int j = (int)(0); j < (int)(V[i].size()); j++) C[i][V[i][j]]--;
    V[i].clear();
  }
  for (int i = (int)(0); i < (int)(W); i++)
    for (int j = (int)(0); j < (int)(B); j++) {
      if (i * B + j >= N) continue;
      int v = A[i * B + j];
      V[i].push_back(v);
      C[i][v]++;
    }
}
int erase(int pos) {
  int vl = 0, vr = 0;
  for (int i = (int)(0); i < (int)(W); i++) {
    vl = vr;
    vr = vr + V[i].size();
    if (vl <= pos && pos < vr) {
      int val = V[i][pos - vl];
      V[i].erase(V[i].begin() + pos - vl);
      C[i][val]--;
      return val;
    }
  }
  assert(false);
}
void insert(int pos, int val) {
  int vl = 0, vr = 0;
  for (int i = (int)(0); i < (int)(W); i++) {
    vl = vr;
    vr = vr + V[i].size();
    if (vl <= pos && pos < vr) {
      V[i].insert(V[i].begin() + pos - vl, val);
      C[i][val]++;
      return;
    } else if (pos == vr) {
      V[i].push_back(val);
      C[i][val]++;
      return;
    }
  }
  assert(false);
}
int query(int l, int r, int k) {
  int vl = 0, vr = 0, res = 0;
  for (int i = (int)(0); i < (int)(W); i++) {
    vl = vr;
    vr = vr + V[i].size();
    if (vr <= l || r <= vl) continue;
    if (l <= vl && vr <= r) {
      res += C[i][k];
    } else {
      int index = vl;
      for (int j = (int)(0); j < (int)(V[i].size()); j++) {
        if (l <= index && index < r && V[i][j] == k) res++;
        index++;
      }
    }
  }
  return res;
}
int main() {
  int N, Q, L = 0;
  scanf("%d", &N);
  V[0].resize(N);
  for (int i = (int)(0); i < (int)(N); i++) {
    scanf("%d", &V[0][i]);
    C[0][V[0][i]]++;
  }
  scanf("%d", &Q);
  for (int q = (int)(0); q < (int)(Q); q++) {
    if (q % B == 0) re_build(N);
    int t, l, r, k;
    scanf("%d", &t);
    if (t == 1) {
      scanf("%d%d", &l, &r);
      l = (l + L - 1) % N;
      r = (r + L - 1) % N;
      if (r < l) swap(l, r);
      int v = erase(r);
      insert(l, v);
    } else {
      scanf("%d%d%d", &l, &r, &k);
      l = (l + L - 1) % N;
      r = (r + L - 1) % N;
      k = (k + L - 1) % N + 1;
      if (r < l) swap(l, r);
      int res = L = query(l, r + 1, k);
      printf("%d\n", res);
    }
  }
  return 0;
}