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
#include <bits/stdc++.h>
using namespace std;
const int NMAX = 100000 + 5;
const int RAD = 440;
int whichBucket[NMAX];
int bucketLeft[NMAX];
int bucketRight[NMAX];
int buckets;
deque<int> dqs[(NMAX + RAD - 1) / RAD + 1];
int Maps[(NMAX + RAD - 1) / RAD + 1][NMAX];
int N, Q;
int a[NMAX];
void precompute() {
  for (int i = 1; i <= N; i += RAD) {
    bucketLeft[++buckets] = i;
    for (int j = i; j <= N && j < i + RAD; ++j) {
      bucketRight[buckets] = j;
      whichBucket[j] = buckets;
      dqs[buckets].push_back(a[j]);
      ++Maps[buckets][a[j]];
    }
  }
}
int count(int val, int l, int r) {
  int lb = whichBucket[l];
  int rb = whichBucket[r];
  int ans = 0;
  if (lb == rb) {
    int start = bucketLeft[lb];
    for (int i = l; i <= r; ++i) ans += (dqs[lb][i - start] == val);
  } else {
    int start = bucketLeft[lb];
    for (int i = l; i <= bucketRight[lb]; ++i)
      ans += (dqs[lb][i - start] == val);
    start = bucketLeft[rb];
    for (int i = start; i <= r; ++i) ans += (dqs[rb][i - start] == val);
    for (int b = lb + 1; b < rb; ++b) ans += Maps[b][val];
  }
  return ans;
}
void shift(int l, int r) {
  int lb = whichBucket[l];
  int rb = whichBucket[r];
  if (lb == rb) {
    int start = bucketLeft[lb];
    int val = dqs[lb][r - start];
    dqs[lb].erase(dqs[lb].begin() + (r - start));
    dqs[lb].insert(dqs[lb].begin() + (l - start), val);
  } else {
    int start = bucketLeft[rb];
    int val = dqs[rb][r - start];
    dqs[rb].erase(dqs[rb].begin() + (r - start));
    Maps[rb][val]--;
    start = bucketLeft[lb];
    dqs[lb].insert(dqs[lb].begin() + (l - start), val);
    Maps[lb][val]++;
    for (int b = lb; b < rb; ++b) {
      val = dqs[b].back();
      dqs[b].pop_back();
      Maps[b][val]--;
      dqs[b + 1].push_front(val);
      Maps[b + 1][val]++;
    }
  }
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> N;
  for (int i = 1; i <= N; ++i) cin >> a[i];
  precompute();
  cin >> Q;
  int lastAns = 0;
  while (Q--) {
    int type, l, r;
    cin >> type >> l >> r;
    l = (l + lastAns - 1) % N + 1;
    r = (r + lastAns - 1) % N + 1;
    if (r < l) swap(l, r);
    if (type == 1)
      shift(l, r);
    else {
      int k;
      cin >> k;
      k = (k + lastAns - 1) % N + 1;
      lastAns = count(k, l, r);
      cout << lastAns << '\n';
    }
  }
  return 0;
}
