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.*;

class Main{
  public static void main(String[] args){
    Scanner sc=new Scanner(System.in);
    int n=sc.nextInt();
    String s=sc.next();
    NavigableSet<Integer>[] nsa=new NavigableSet[26];
    for(int i=0;i<26;i++){
      nsa[i]=new TreeSet<Integer>();
    }
    for(int i=0;i<n;i++){
      nsa[s.charAt(i)-'a'].add(i);
    }
    int q=sc.nextInt();
    for(int i=0;i<q;i++){
      if(sc.nextInt()==1){
        int idx=sc.nextInt();
        int ch=sc.next().charAt(0)-'a';
        for(int j=0;j<26;j++){
          if(ch==j){
            nsa[j].add(idx-1);
          }else{
            nsa[j].remove(idx-1);
          }
        }
      }else{
        int l=sc.nextInt()-1;
        int r=sc.nextInt()-1;
        int ans=0;
        for(int j=0;j<26;j++){
          Integer e=nsa[j].ceiling(l);
          if(e!=null&&e<=r){
            ans++;
          }
        }
        System.out.println(ans);
      }
    }
  }
}