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 JAVA):

Solution:
import java.util.*;
import java.lang.*;
import java.io.*;

public class D
{
	private static final int maxSize = 100000 + 13;
	private static final int blockSize = 350;
	private static final int blockCount = maxSize/blockSize + 13;

	private static int [] a = new int[maxSize];

	private static int currBlock;
	private static int [][] values = new int[blockCount][2*blockSize + 13];
	private static int [] valBlockSize = new int[blockCount];
	private static int [][] stored = new int[blockCount][2*blockSize + 13];
	private static int [] storedBlockSize = new int[blockCount];
	private static short [][] cnt = new short[blockCount][maxSize];

	private static int arraySize;

	public static void main(String [] args) throws IOException
	{
		BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter writer = new PrintWriter(new OutputStreamWriter(System.out));

		arraySize = Integer.parseInt(reader.readLine());
		StringTokenizer tokenizer = new StringTokenizer(reader.readLine());

		for(int i = 0 ; i < arraySize ; i++)
			a[i] = Integer.parseInt(tokenizer.nextToken());

		build();

		int q = Integer.parseInt(reader.readLine());
		int lastAns = 0;
		int c = 1;

		while(q-- > 0)
		{
			tokenizer = new StringTokenizer(reader.readLine());
			int type = Integer.parseInt(tokenizer.nextToken());
			int lt = Integer.parseInt(tokenizer.nextToken());
			int rt = Integer.parseInt(tokenizer.nextToken());

			int l = (lt+lastAns-1)%arraySize;
			int r = (rt+lastAns-1)%arraySize;

			if(l > r)
			{
				int t = l;
				l = r;
				r = t;
			}

			if(type == 1)
			{
				shift(l, r);
				c++;
			}
			else
			{
				int kt = Integer.parseInt(tokenizer.nextToken());
				int k = (kt+lastAns-1)%arraySize + 1;

				int res = count(l, r, k);
				lastAns = res;

				writer.println(res);
			}

			if(c%blockSize == 0)
			{
				c++;
				reBuild();
			}
		}

		writer.flush();
		writer.close();

	}

	private static void build()
	{
		currBlock = 0;
		for(int i = 0 ; i < arraySize ; i += blockSize, currBlock++)
			for(int j = i ; j < Math.min(arraySize, i+blockSize) ; j++)
			{
				values[currBlock][valBlockSize[currBlock]++] = a[j];
				stored[currBlock][storedBlockSize[currBlock]++] = a[j];
				cnt[currBlock][a[j]]++;
			}
	}

	private static void reBuild()
	{
		int at = 0;

		for(int i = 0 ; at < arraySize ; i++)
			for(int j = 0 ; j < valBlockSize[i] ; j++)
				a[at++] = values[i][j];

		for(int i = 0 ; i < blockCount ; i++)
		{
			for(int j = 0 ; j < valBlockSize[i] ; j++)
				values[i][j] = 0;
			for(int j = 0 ; j < storedBlockSize[i] ; j++)
			{
				cnt[i][stored[i][j]] = 0;
				stored[i][j] = 0;
			}

			valBlockSize[i] = storedBlockSize[i] = 0;
		}

		build();
	}

	private static int count(int l, int r, int k)
	{
		return count(r, k) - count(l-1, k);
	}

	private static int count(int r, int k)
	{
		if(r < 0)
			return 0;

		int res = 0;
		int sum = 0;
		for(int i = 0 ; i < currBlock ; i++)
		{
			sum += valBlockSize[i];
			if(sum <= r+1)
				res += cnt[i][k];
			else
			{
				sum -= valBlockSize[i];
				for(int j = 0 ; sum <= r ; j++, sum++)
					if(values[i][j] == k)
						res++;

				break;
			}
		}

		return res;
	}

	private static void shift(int l, int r)
	{
		if(l == r)
			return;
		int x = erase(r);
		insert(l, x);
	}

	private static void insert(int l, int x)
	{
		int sum = 0;
		for(int i = 0 ; ; i++)
		{
			sum += valBlockSize[i];
			if(sum > l)
			{
				stored[i][storedBlockSize[i]++] = x;
				cnt[i][x]++;
				int pos = l - (sum-valBlockSize[i]);

				for(int j = valBlockSize[i] ; j > pos ; j--)
					values[i][j] = values[i][j-1];
				values[i][pos] = x;
				valBlockSize[i]++;

				return;
			}
		}
	}

	private static int erase(int r)
	{
		int sum = 0;
		for(int i = 0 ; ; i++)
		{
			sum += valBlockSize[i];
			if(sum > r)
			{
				int pos = r - (sum-valBlockSize[i]);
				cnt[i][values[i][pos]]--;
				int res = values[i][pos];
				valBlockSize[i]--;

				for(int j = pos ; j < valBlockSize[i] ; j++)
					values[i][j] = values[i][j+1];

				return res;
			}
		}
	}
}