You are given a string S of length N consisting of lowercase English letters.

Process Q queries of the following two types:

* Type 1: change the i_q-th character of S to c_q. (Do nothing if the i_q-th character is already c_q.)
* Type 2: answer the number of different characters occurring in the substring of S between the l_q-th and r_q-th characters (inclusive).

Constraints

* N, Q, i_q, l_q, and r_q are integers.
* S is a string consisting of lowercase English letters.
* c_q is a lowercase English letter.
* 1 \leq N \leq 500000
* 1 \leq Q \leq 20000
* |S| = N
* 1 \leq i_q \leq N
* 1 \leq l_q \leq r_q \leq N
* There is at least one query of type 2 in each testcase.

Input

Input is given from Standard Input in the following format:


N
S
Q
Query_1
\vdots
Query_Q


Here, Query_i in the 4-th through (Q+3)-th lines is one of the following:


1 i_q c_q



2 l_q r_q


Output

For each query of type 2, print a line containing the answer.

Example

Input

7
abcdbbd
6
2 3 6
1 5 z
2 1 1
1 4 a
1 7 d
2 1 7


Output

3
1
5
import java.util.*;
import java.io.PrintWriter;
public class Main{
	public static void main(String[] args){
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt();
		char[] s=sc.next().toCharArray();
		int q=sc.nextInt();
		TreeSet<Integer>[] set=new TreeSet[26];
		for(int i=0; i<26; i++){
			set[i]=new TreeSet<>();
		}
		for(int i=0; i<n; i++){
			set[s[i]-'a'].add(i);
		}
		for(int i=0; i<q; i++){
			int a=sc.nextInt();
			if(a==1){
				int b=sc.nextInt()-1;
				char c=sc.next().charAt(0);
				set[s[b]-'a'].remove(b);
				set[c-'a'].add(b);
				s[b]=c;
			}else{
				int b=sc.nextInt()-1;
				int c=sc.nextInt()-1;
				int ans=0;
				for(int j=0; j<26; j++){
					 Integer ch=set[j].ceiling(b);
         			 if(ch!=null&&ch<=c){
            			ans++;
          			}
				}
				System.out.println(ans);
			}
		}
	}
}
