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) throws Exception {
        // Your code here!
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        String str = sc.next();
        char c[] = str.toCharArray();
        
        // set: 英小文字の出現位置
        TreeSet[] set = new TreeSet[26];
        for (int i = 0; i < 26; i++) {
            set[i] = new TreeSet<Integer>();
        }
        for (int i = 0; i < n; i++) {
            char chr = c[i];
            set[chr - 'a'].add(i);
        }
        
        int query = sc.nextInt();
        while (query-- > 0) {
            int type = sc.nextInt();
            
            if (type == 1) {
                // abcb => azcb
                //  ^       ^
                
                // chrの出現位置を消し、chr2の出現位置を追加、c[]を書き換え
                int index = sc.nextInt() - 1;
                char chr = c[index];
                char chr2 = (sc.next()).charAt(0);
                set[chr - 'a'].remove(index);
                set[chr2 - 'a'].add(index);
                c[index] = chr2;
            } else {
                // abcdbbd => 3
                //  ^^^^^
                
                // start以上end以下区間内にある英小文字の種類数を出力
                int start = sc.nextInt() - 1;
                int end = sc.nextInt() - 1;
                int count = 0;
                for (int i = 0; i < 26; i++) {
                    if (set[i].ceiling(start) == null) continue;
                    
                    if ((int)set[i].ceiling(start) <= end) {
                        count++;
                    }
                }
                System.out.println(count);
            }
        }
        
    }
}