You are given a string s consisting of lowercase Latin letters and q queries for this string.

Recall that the substring s[l; r] of the string s is the string s_l s_{l + 1} ... s_r. For example, the substrings of "codeforces" are "code", "force", "f", "for", but not "coder" and "top".

There are two types of queries: 

  * 1~ pos~ c (1 ≤ pos ≤ |s|, c is lowercase Latin letter): replace s_{pos} with c (set s_{pos} := c); 
  * 2~ l~ r (1 ≤ l ≤ r ≤ |s|): calculate the number of distinct characters in the substring s[l; r]. 

Input

The first line of the input contains one string s consisting of no more than 10^5 lowercase Latin letters.

The second line of the input contains one integer q (1 ≤ q ≤ 10^5) — the number of queries.

The next q lines contain queries, one per line. Each query is given in the format described in the problem statement. It is guaranteed that there is at least one query of the second type.

Output

For each query of the second type print the answer for it — the number of distinct characters in the required substring in this query.

Examples

Input


abacaba
5
2 1 4
1 4 b
1 5 b
2 4 6
2 1 7


Output


3
1
2


Input


dfcbbcfeeedbaea
15
1 6 e
1 4 b
2 6 14
1 7 b
1 12 c
2 6 8
2 1 6
1 7 c
1 2 f
1 10 a
2 7 9
1 10 a
1 14 b
1 1 f
2 1 11


Output


5
2
5
2
6
import java.io.*;
import java.util.*;

public class Main{
	public static class FastReader {
		BufferedReader br;
		StringTokenizer st;
		//it reads the data about the specified Pair and divide the data about it ,it is quite fast
		//than using direct 
 
		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
 
		String next() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (Exception r) {
					r.printStackTrace();
				}
			}
			return st.nextToken();
		}
 
		int nextInt() {
			return Integer.parseInt(next());//converts string to integer
		}
 
		double nextDouble() {
			return Double.parseDouble(next());
		}
 
		long nextLong() {
			return Long.parseLong(next());
		}
 
		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (Exception r) {
				r.printStackTrace();
			}
			return str;
		}
	}
	
	public static PrintWriter out = new PrintWriter (new BufferedOutputStream(System.out));
	static long mod = (long) (1e9+7);
	
	public static void main(String[] args) {
		FastReader sc = new FastReader();
		 char[] s = sc.next().toCharArray();
	        int q = sc.nextInt();
	        TreeMap<Integer,Boolean>[] map = new TreeMap[26];
	        for(int i=0;i<26;++i) map[i] = new TreeMap<>();
	        for(int i=0;i<s.length;++i){
	            map[s[i]-'a'].put(i+1,true);
	        }
	        while(q-->0){
	         int type = sc.nextInt();
	         if(type == 1){
	             int pos = sc.nextInt()-1;
	             char c = sc.next().charAt(0);
	             char prev = s[pos];
	             if(map[prev-'a'].containsKey(pos+1)) map[prev-'a'].remove(pos+1);
	             map[c-'a'].put(pos+1,true);
	             s[pos] = c;
	         }else{
	             int l = sc.nextInt();
	             int r = sc.nextInt();
	             int ans = 0;
	             for(int i=0;i<26;++i){
	                 if(map[i].ceilingKey(l) != null){
	                     int key = map[i].ceilingKey(l);
	                     
	                     if(key<=r) 	
	                    	 ++ans;
	                     
	                 }

	             }
	             out.println(ans);
	            
	         }
              
	        }
	    out.close();
	}

	
	
}
