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.math.BigInteger;
import java.util.*;
import java.util.Map.Entry;
 
 
public class  gym{
	static int INF=(int)2e9;
	static class ftree{
		int n;
		long[]ft;
		ftree(int z){
			ft=new long[z+1];
			n=z;
		}
		void updatesum(int idx,int k) {
			while(idx<=n) {
				ft[idx]+=k;
				idx+=(idx&(-1*idx));
			}
			
		}
		int querysum(int idx) {
			int sum=0;
			while(idx>0) {
				sum+=ft[idx];
				idx-=(idx&(-1*idx));
			}
			return sum;
		}
		
	}
	public static void main(String[] args) throws Exception{
    	MScanner sc = new MScanner(System.in);
        PrintWriter pw=new PrintWriter(System.out);
        char[]in=sc.nextLine().toCharArray();
        int q=sc.nextInt();
        int n = in.length;
		ftree[]le=new ftree[26];
		for(int i=0;i<26;i++) {
			le[i]=new ftree(n+2);
		}
		for(int i=0;i<n;i++) {
			le[in[i]-'a'].updatesum(i+1, 1);
		}
		while(q-->0) {
			if(sc.nextInt()==1) {
				int pos=sc.nextInt()-1;
				int prev=(in[pos]-'a');
				in[pos]=sc.nextChar();
				int newc=(in[pos]-'a');
				le[prev].updatesum(pos+1, -1);
				le[newc].updatesum(pos+1, 1);
			}
			else {
				int l=sc.nextInt(),r=sc.nextInt();
				int c=0;
				for(int i=0;i<26;i++) {
					int x=le[i].querysum(r)-le[i].querysum(l-1);
					//System.out.println(x);
					if(x>0)c++;
				}
				pw.println(c);
			}
		}
        pw.flush();
    }
    static class MScanner {
		StringTokenizer st;
		BufferedReader br;
		public MScanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}
 
		public MScanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}
 
		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}
		public int[] takearr(int n) throws IOException {
	        int[]in=new int[n];for(int i=0;i<n;i++)in[i]=nextInt();
	        return in;
		}
		public String nextLine() throws IOException {
			return br.readLine();
		}
 
		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}
 
		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}
 
		public char nextChar() throws IOException {
			return next().charAt(0);
		}
 
		public Long nextLong() throws IOException {
			return Long.parseLong(next());
		}
 
		public boolean ready() throws IOException {
			return br.ready();
		}
 
		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
	}
}