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 Dangerous {
    private static final int maxSize = 100000 + 13;//size of array
    private static final int blockSize = 350;//sqrt(n)
    private static final int blockCount = maxSize/blockSize + 13;
    private static int [] a = new int[maxSize];//array
    private static int currBlock;//block count after build
    private static int [][] values = new int[blockCount][2*blockSize + 13];
    private static int [] valBlockSize = 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];
                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++) {
                cnt[i][values[i][j]] = 0;
                values[i][j] = 0;
            }
            valBlockSize[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) {
                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;
            }
        }
    }
}