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 N = 101010;
int const sq = 320;
int n, a[N];
int cnt[sq][N];
deque<int> v[sq];
void build() {
  for (int i = 0; i <= n; i++) {
    int j = i / sq;
    cnt[j][a[i]]++;
    v[j].push_back(a[i]);
  }
}
void shift(int l, int r) {
  int aL = l / sq;
  int aR = r / sq;
  int il = l % sq;
  int ir = r % sq;
  if (aL == aR) {
    vector<int> cur;
    while (v[aL].size()) {
      cur.push_back(v[aL].front());
      v[aL].pop_front();
    }
    for (int i = 0; i < il; i++) v[aL].push_back(cur[i]);
    v[aL].push_back(cur[ir]);
    for (int i = il; i < ir; i++) v[aL].push_back(cur[i]);
    for (int i = ir + 1; i < cur.size(); i++) v[aL].push_back(cur[i]);
    return;
  }
  vector<int> curL, curR;
  while (v[aL].size()) {
    curL.push_back(v[aL].front());
    v[aL].pop_front();
  }
  while (v[aR].size()) {
    curR.push_back(v[aR].front());
    v[aR].pop_front();
  }
  for (int i = 0; i < il; i++) v[aL].push_back(curL[i]);
  v[aL].push_back(curR[ir]);
  cnt[aL][curR[ir]]++;
  for (int i = il; i + 1 < curL.size(); i++) v[aL].push_back(curL[i]);
  cnt[aL][curL.back()]--;
  int last = curL.back();
  for (int i = aL + 1; i < aR; i++) {
    cnt[i][last]++;
    v[i].push_front(last);
    last = v[i].back();
    cnt[i][last]--;
    v[i].pop_back();
  }
  cnt[aR][last]++;
  v[aR].push_back(last);
  for (int i = 0; i < ir; i++) v[aR].push_back(curR[i]);
  cnt[aR][curR[ir]]--;
  for (int i = ir + 1; i < curR.size(); i++) v[aR].push_back(curR[i]);
}
int ans;
int act(int c) { return ((c + ans - 1) % n) + 1; }
int query(int l, int r, int val) {
  int aL = l / sq;
  int aR = r / sq;
  int il = l % sq;
  int ir = r % sq;
  int res = 0;
  if (aL == aR) {
    vector<int> cur;
    while (v[aL].size()) {
      cur.push_back(v[aL].front());
      v[aL].pop_front();
    }
    for (int i = 0; i < il; i++) v[aL].push_back(cur[i]);
    for (int i = il; i <= ir; i++) {
      v[aL].push_back(cur[i]);
      if (cur[i] == val) res++;
    }
    for (int i = ir + 1; i < cur.size(); i++) v[aL].push_back(cur[i]);
    return res;
  }
  vector<int> curL, curR;
  while (v[aL].size()) {
    curL.push_back(v[aL].front());
    v[aL].pop_front();
  }
  while (v[aR].size()) {
    curR.push_back(v[aR].front());
    v[aR].pop_front();
  }
  for (int i = 0; i < il; i++) v[aL].push_back(curL[i]);
  for (int i = il; i < curL.size(); i++) {
    v[aL].push_back(curL[i]);
    if (curL[i] == val) res++;
  }
  for (int i = aL + 1; i < aR; i++) {
    res += cnt[i][val];
  }
  for (int i = 0; i <= ir; i++) {
    v[aR].push_back(curR[i]);
    if (curR[i] == val) res++;
  }
  for (int i = ir + 1; i < curR.size(); i++) v[aR].push_back(curR[i]);
  return res;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  build();
  int m;
  cin >> m;
  while (m--) {
    int t;
    scanf("%d", &t);
    if (t == 1) {
      int l, r;
      scanf("%d%d", &l, &r);
      l = act(l);
      r = act(r);
      if (l > r) swap(l, r);
      shift(l, r);
    } else {
      int l, r, k;
      scanf("%d%d%d", &l, &r, &k);
      l = act(l);
      r = act(r);
      k = act(k);
      if (l > r) swap(l, r);
      printf("%d\n", ans = query(l, r, k));
    }
  }
}