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.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class Round590D {

	public static class node{
		int[] arr;
		public node() {
			arr=new int[26];
		}
	}
	
	public static node combine(node x,node y) {
		node ans =new node();
		for(int i=0;i<26;i++) {
			ans.arr[i]=x.arr[i]+y.arr[i];
		}
		return ans;
	}
	
	public static void Build(node[] tree,int start, int end,int index,String str) {
		if(start==end) {
			tree[index]=new node();
			tree[index].arr[str.charAt(start)-'a']=1;
			return;
		}
		int mid = (start+end)>>1;
		Build(tree,start,mid,index*2,str);
		Build(tree,mid+1,end,index*2+1,str);
		tree[index]=combine(tree[index*2],tree[index*2+1]);
	}
	
	public static void Update(node[] tree,int start,int end,int index,int i,char val) {
		if(i<start || i>end) {
			return;
		}
		if(start==end && start == i) {
			tree[index]=new node();
			tree[index].arr[val-'a']=1;
			return;
		}
		int mid = (start+end)>>1;
		Update(tree,start,mid,index*2,i,val);
		Update(tree,mid+1,end,index*2+1,i,val);
		tree[index]=combine(tree[index*2],tree[index*2+1]);
	}
	
	public static node Query(node[] tree,int start,int end,int index,int l,int r) {
		if(r<start || l>end) {
			return new node();
		}
		if(l<=start && r>=end) {
			return tree[index];
		}
		int mid = (start + end) >> 1;
		node left = Query (tree, start , mid , index*2,l,r);
		node right = Query(tree,mid+1,end,index*2+1,l,r);
		return combine(left,right);
	}
	
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		out=new PrintWriter(new BufferedOutputStream(System.out));
		FastReader s=new FastReader();
		String str=s.next();
		int n=str.length();
		node[] tree=new node[4*n+1];
		Build(tree, 0, n-1, 1, str);
		int q=s.nextInt();
		while(q-->0) {
			int type=s.nextInt();
			if(type==1) {
				int i=s.nextInt()-1;
				char val=s.next().charAt(0);
				Update(tree, 0, n-1, 1, i, val);
			}else {
				int l=s.nextInt()-1;
				int r=s.nextInt()-1;
				node ans = Query(tree, 0, n-1, 1, l, r);
				int count=0;
				for(int i=0;i<26;i++) {
					if(ans.arr[i]!=0)count++;
				}
				out.println(count);
			}
		}
		out.close();
	}
	
	public static PrintWriter out;
	
	public static class FastReader {
		BufferedReader br;
		StringTokenizer st;
		//it reads the data about the specified point and divide the data about it ,it is quite fast
		//than using direct 

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
			//InputStream reads the data and decodes in character stream
			//It acts as bridge between byte stream and character stream
		}

		String next() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (Exception e) {
					e.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 e) {
				e.printStackTrace();
			}
			return str;
		}
	}

}
