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[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		char[] s = sc.next().toCharArray();
		int Q = sc.nextInt();
		TreeSet<Integer>[] t = new TreeSet[26];
		for(int i = 0; i < 26; i++)
			t[i] = new TreeSet<Integer>();
		for(int i = 0; i < N; i++)
			t[s[i]-'a'].add(i);
		for(int i = 0; i < Q; i++){
			if(sc.nextInt() == 1) {
				int q = sc.nextInt()-1;
				char c = sc.next().charAt(0);
				t[s[q]-'a'].remove(q);
				t[c-'a'].add(q);
				s[q] = c;
			}
			else {
				int ans = 0;			
				int l =  sc.nextInt()-1, r = sc.nextInt()-1;
				for(char j = 0; j < 26; j++)
						if(t[j].ceiling(l) != null && t[j].ceiling(l) <= r)
							ans++;
				System.out.println(ans);
			}
		}

	}

}