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.*;

public class Main{
	public static void main(String[] $){
		Scanner s=new Scanner(System.in);
		int n=s.nextInt();
		char[]t=s.next().toCharArray();

		ArrayList<TreeSet<Integer>>m=new ArrayList<>(26);
		for(int i=0;i<26;++i)
			m.add(new TreeSet<>());
		for(int i=0;i<n;++i)
			m.get(t[i]-'a').add(i);

		for(int q=s.nextInt();q>0;--q) {
			if(s.nextInt()==1) {
				int a=s.nextInt()-1;
				char b=s.next().charAt(0);
				if(t[a]!=b) {
					m.get(t[a]-'a').remove(a);
					m.get(b-'a').add(a);
					t[a]=b;
				}
			}else {
				int a=s.nextInt();
				int b=s.nextInt();
				int c=0;
				for(TreeSet<Integer> o:m) {
					if(!o.subSet(a-1,b).isEmpty())
						++c;
				}
				System.out.println(c);
			}
		}
	}
}