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();

        TreeSet<Integer>[] sets = new TreeSet[26];
        for(int i=0; i<26; i++) {
            sets[i] = new TreeSet<>();
        }

        char[] s = sc.next().toCharArray();
        int[] c = new int[n];
        for(int i=0; i<n; i++) {
            c[i] = s[i]-'a';
            sets[c[i]].add(i);
        }

        int q = sc.nextInt();
        for(int i=0; i<q; i++) {
            int com = sc.nextInt();
            if(com == 1) {
                int ind = sc.nextInt()-1;
                int num = sc.next().charAt(0) - 'a';
                if(c[ind] != num) {
                    sets[c[ind]].remove(ind);
                    sets[num].add(ind);
                    c[ind] = num;
                }
            } else {
                int l = sc.nextInt()-1;
                int r = sc.nextInt()-1;
                int sum = 0;
                for(int j=0; j<26; j++) {
                    Integer h = sets[j].higher(l-1);
                    if(h != null && h<=r) sum++;
                }
                System.out.println(sum);

            }

        }

    }

}