Problem: You are given a string s of length n, consisting of first k lowercase English letters.

We define a c-repeat of some string q as a string, consisting of c copies of the string q. For example, string "acbacbacbacb" is a 4-repeat of the string "acb".

Let's say that string a contains string b as a subsequence, if string b can be obtained from a by erasing some symbols.

Let p be a string that represents some permutation of the first k lowercase English letters. We define function d(p) as the smallest integer such that a d(p)-repeat of the string p contains string s as a subsequence.

There are m operations of one of two types that can be applied to string s:

  1. Replace all characters at positions from li to ri by a character ci. 
  2. For the given p, that is a permutation of first k lowercase English letters, find the value of function d(p). 



All operations are performed sequentially, in the order they appear in the input. Your task is to determine the values of function d(p) for all operations of the second type.

Input

The first line contains three positive integers n, m and k (1 ≤ n ≤ 200 000, 1 ≤ m ≤ 20000, 1 ≤ k ≤ 10) — the length of the string s, the number of operations and the size of the alphabet respectively. The second line contains the string s itself.

Each of the following lines m contains a description of some operation: 

  1. Operation of the first type starts with 1 followed by a triple li, ri and ci, that denotes replacement of all characters at positions from li to ri by character ci (1 ≤ li ≤ ri ≤ n, ci is one of the first k lowercase English letters). 
  2. Operation of the second type starts with 2 followed by a permutation of the first k lowercase English letters.

Output

For each query of the second type the value of function d(p).

Examples

Input

7 4 3
abacaba
1 3 5 b
2 abc
1 4 4 c
2 cba


Output

6
5

Note

After the first operation the string s will be abbbbba.

In the second operation the answer is 6-repeat of abc: ABcaBcaBcaBcaBcAbc.

After the third operation the string s will be abbcbba.

In the fourth operation the answer is 5-repeat of cba: cbAcBacBaCBacBA.

Uppercase letters means the occurrences of symbols from the string s.

Solution (in JAVA):

Solution:
/**
 * Created by Aminul on 8/23/2017.
 */

import java.io.*;
import java.util.*;

public class CF610E {
    public static int [][]cnt;
    public static void main(String[] args)throws Exception {
        InputReader in = new InputReader(System.in);
        PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out));
        cnt = new int[200][200];
        int n = in.nextInt(), q = in.nextInt(), k = in.nextInt();
        char [] s = in.next().toCharArray();
        Interval_Map imap = new Interval_Map();
        imap.build(n, s);
        for(int i = 0; i < q; i++){
            int type = in.nextInt();
            if(type == 1){
                int l = in.nextInt() , r = in.nextInt();
                char c = in.next().charAt(0);
                imap.range_update(l, r, c);
            }
            else{
                char [] t = in.next().toCharArray();
                int ans = n;
                for(int a = 0; a < k; a++){
                    for(int b = a+1; b < k; b++){
                        ans -= cnt[t[a]][t[b]];
                    }
                }

                pw.println(ans);
            }
        }




        pw.close();
    }

    static class Interval_Map{
        TreeMap<Integer, Character> map;
        Interval_Map(){
            map = new TreeMap<>();
        }

        void build(int n, char[]s){
            map.put(0, '0');
            map.put(n+1, '1');
            for(int i = 0; i < n; i++){
                map.put(i+1, s[i]);
                if(i+1 < n) cnt[s[i]][s[i+1]]++;
            }
        }

        void range_update(int l, int r, char c){
            int left = map.lowerKey(l);
            int right = map.floorKey(r), right2 = map.higherKey(r);
            if(r+1 < right2){
                map.put(r+1, map.get(right));
                right2 = r+1;
            }

            int key1 = left;
            for(int key2: map.subMap(left, false, right2, true).keySet()){
                cnt[map.get(key1)][map.get(key2)]--;
                key1 = key2;
            }

            map.subMap(left, false, right2, false).clear();

            map.put(l, c);
            cnt[map.get(left)][c]++;
            cnt[c][map.get(right2)]++;
        }

    }

    static void debug(Object...obj) {
        System.err.println(Arrays.deepToString(obj));
    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream));
            tokenizer = null;
        }

        public String next()throws Exception {
            while (tokenizer == null || !tokenizer.hasMoreTokens())
                tokenizer = new StringTokenizer(reader.readLine());
            return tokenizer.nextToken();
        }

        public String nextLine()throws Exception {
            String line = null;
            tokenizer = null;
            line =  reader.readLine();
            return line;
        }

        public int nextInt()throws Exception {
            return Integer.parseInt(next());
        }

        public double nextDouble() throws Exception{
            return Double.parseDouble(next());
        }

        public long nextLong()throws Exception {
            return Long.parseLong(next());
        }

    }
}