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();
        String S = sc.next();
        char[] sca = S.toCharArray();
        List<BIT> cnts = new ArrayList<>();
        for(int i=0;i<26;i++){
            cnts.add(new BIT(N+10));
        }
        for(int i=0;i<N;i++){
            int ind = S.charAt(i)-'a';
            cnts.get(ind).addTo(i + 1, 1);
        }
        int Q = sc.nextInt();
        for(int i=0;i<Q;i++){
            int qt = sc.nextInt();
            if(qt == 1){
                int iq=sc.nextInt()-1;
                String cq=sc.next();
                int toc=cq.charAt(0)-'a';
                int fromc = sca[iq]-'a';
                if(fromc != toc) {
                    cnts.get(fromc).addTo(iq + 1, -1);
                    sca[iq] = cq.charAt(0);
                    cnts.get(toc).addTo(iq + 1, 1);
                }
            }else{
                int lq=sc.nextInt();
                int rq=sc.nextInt();
                long res = cnts.stream().filter(bit->
                        bit.sumTo(rq)-bit.sumTo(lq-1)>0
                ).count();
                System.out.println(res);
            }
        }


    }
    private static class BIT{
        int[] vals;
        int len;
        BIT(int n){
            len = n;
            vals = new int[len+1];
            Arrays.fill(vals, 0);
        }
        public void addTo(int ind, int val){
            while(ind <= len){
                vals[ind] += val;
                ind += btmBit(ind);
            }
        }

        public int sumTo(int ind){
            int sum = 0;
            while(ind > 0){
                sum += vals[ind];
                ind -= btmBit(ind);
            }
            return sum;
        }

        private static int btmBit(int a){
            return a&(-a);
        }

        @Override
        public String toString(){
            StringBuilder bld = new StringBuilder();
            for(int i=0;i<len+1;i++){
                bld.append(sumTo(i));
            }
            return bld.toString();
        }
    }
}
