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.util.BitSet;
import java.util.HashSet;
import java.util.Scanner;
import java.util.Set;

public class Main {
    static Node[] nodes = new Node[101000];
    static class Node {
//        Set<Character> unique = new HashSet<>();
        BitSet unique = new BitSet(26);
        char leaf;
        int start;
        int end;
        public Node left;
        public Node right;
        public Node parent;
    }

    //Build the segment tree
    static Node build(char[] v, int start, int end)
    {
        Node root = new Node();
        if(start == end)
        {
            root.unique.set(v[start]-'a');
            root.leaf = v[start];
            nodes[start] = root;
        }
        else
        {
            for(int i = start; i <= end; i++)
            {
                root.unique.set(v[i]-'a');
            }
            root.left = build(v, start, start + (end - start)/2);
            root.left.parent = root;
            root.right = build(v, start + (end - start)/2 + 1, end);
            root.right.parent = root;
        }
        root.start = start;
        root.end = end;
        return root;
    }
    static BitSet findRange(Node root, int start, int end)
    {
        BitSet lProduct = new BitSet(26);
        BitSet rProduct = new BitSet(26);
        BitSet output = new BitSet(26);
        if(root.right == null && root.left == null) return root.unique;
        if(root.start >= start && root.end <= end) return root.unique;
        if(start < root.right.start)
        {
            lProduct = findRange(root.left, start, end);
        }
        if(end > root.left.end)
        {
            rProduct = findRange(root.right, start, end);
        }
        output.or(rProduct);
        output.or(lProduct);
        return output;
    }
    //Start at the node at the bottom of the tree and work upwards to the root of the tree, updating the products of the parent nodes along the way.
    static void update(int index, char newChar)
    {
        Node actualNode = nodes[index];
        int oldIndex = actualNode.leaf - 'a';
        actualNode.leaf = newChar;
        int newIndex = newChar - 'a';
        if(!actualNode.unique.get(newIndex))
        {
            actualNode.unique.clear(oldIndex);
            actualNode.unique.set(newIndex);
            Node currentNode = actualNode.parent;
            while(currentNode != null)
            {
                if(currentNode.left.unique.get(oldIndex) || currentNode.right.unique.get(oldIndex))
                {
                    if(currentNode.unique.get(newIndex)) break;
                    else
                    {
                        currentNode.unique.set(newIndex);
                        currentNode = currentNode.parent;
                    }
                }
                else
                {
                    currentNode.unique.set(newIndex);
                    currentNode.unique.clear(oldIndex);
                    currentNode = currentNode.parent;
                }
            }
        }
    }
    public static void main(String[] args) {
        Scanner scnr = new Scanner(System.in);
        int K;
        String inputStr = scnr.next();
        K = scnr.nextInt();
        char[] array = inputStr.toCharArray();
        Node root = build(array, 0, array.length-1);
        for(int i = 0; i < K; i++)
        {
            int x, y, com;
            char newChar;
            com = scnr.nextInt();
            if(com == 1)
            {
                x = scnr.nextInt();
                newChar = scnr.next().charAt(0);
                update(x-1,newChar);
            }
            else
            {
                x = scnr.nextInt();
                y = scnr.nextInt();
                System.out.println(findRange(root, x-1, y-1).cardinality());
            }
        }
    }
}
			 				 	 	  	    	  				    	