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;
struct bucketElement {
  int bucketIndex, localPosition, globalPosition, value;
};
int N, A[100000 + 5], freq[317 + 5][100000 + 5];
vector<int> bucket[317 + 5];
void Read() {
  scanf("%d", &N);
  for (int i = 1; i <= N; ++i) {
    int tmp;
    scanf("%d", &tmp);
    bucket[1].push_back(tmp);
  }
}
void Rebuild() {
  int last = 0;
  for (int i = 1; i <= 317; ++i) {
    for (int j = 0; j < bucket[i].size(); ++j) {
      freq[i][bucket[i][j]] = 0;
      ++last;
      A[last] = bucket[i][j];
    }
    bucket[i].clear();
  }
  int lastBuck = 1;
  for (int i = 1; i <= N; ++i) {
    if (bucket[lastBuck].size() == 317) {
      ++lastBuck;
    }
    bucket[lastBuck].push_back(A[i]);
    ++freq[lastBuck][A[i]];
  }
}
bucketElement Find(int pos) {
  bucketElement ans;
  int nElements = 0, currBucket = 1;
  ans.globalPosition = pos;
  while (nElements + bucket[currBucket].size() < pos) {
    nElements += bucket[currBucket].size();
    ++currBucket;
  }
  ans.bucketIndex = currBucket;
  ans.localPosition = pos - nElements - 1;
  ans.value = bucket[currBucket][ans.localPosition];
  return ans;
}
void Cycle(int from, int to) {
  bucketElement whereFrom = Find(from), whereTo = Find(to);
  bucket[whereTo.bucketIndex].erase(bucket[whereTo.bucketIndex].begin() +
                                    whereTo.localPosition);
  bucket[whereFrom.bucketIndex].insert(
      bucket[whereFrom.bucketIndex].begin() + whereFrom.localPosition,
      whereTo.value);
  --freq[whereTo.bucketIndex][whereTo.value];
  ++freq[whereFrom.bucketIndex][whereTo.value];
}
int HowMany(int from, int to, int val) {
  int ans = 0;
  bucketElement whereFrom = Find(from), whereTo = Find(to);
  if (whereFrom.bucketIndex == whereTo.bucketIndex) {
    for (int i = whereFrom.localPosition; i <= whereTo.localPosition; ++i) {
      ans += (bucket[whereFrom.bucketIndex][i] == val);
    }
  } else {
    for (int i = whereFrom.localPosition;
         i < bucket[whereFrom.bucketIndex].size(); ++i) {
      ans += (bucket[whereFrom.bucketIndex][i] == val);
    }
    for (int i = whereFrom.bucketIndex + 1; i < whereTo.bucketIndex; ++i) {
      ans += freq[i][val];
    }
    for (int i = 0; i <= whereTo.localPosition; ++i) {
      ans += (bucket[whereTo.bucketIndex][i] == val);
    }
  }
  return ans;
}
void Solve() {
  int Q, lastAns = 0, l, lTonos, r, rTonos, k, kTonos, type;
  scanf("%d", &Q);
  for (int i = 0; i < Q; ++i) {
    if (i % 317 == 0) {
      Rebuild();
    }
    scanf("%d %d %d", &type, &lTonos, &rTonos);
    l = ((lTonos + lastAns - 1) % N) + 1;
    r = ((rTonos + lastAns - 1) % N) + 1;
    if (r < l) {
      swap(l, r);
    }
    if (type == 1) {
      Cycle(l, r);
    } else if (type == 2) {
      scanf("%d", &kTonos);
      k = ((kTonos + lastAns - 1) % N) + 1;
      lastAns = HowMany(l, r, k);
      printf("%d\n", lastAns);
    }
  }
}
int main() {
  Read();
  Solve();
}