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[] segs = new TreeSet[26];
    	char[] d = sc.next().toCharArray();
    	for (int i = 0; i < 26; i++){
    		segs[i] = new TreeSet();
    		segs[i].add(510000);
    	}
    	for (int i = 0; i < n; i++) {
    		segs[(int) d[i] - 'a'].add(i);
    	}
    	int q = sc.nextInt();
    	for (int i = 0; i < q; i++) {
    		int a = sc.nextInt();
    		if (a==1) {
    			int index = sc.nextInt()-1;
    			char oldc = d[index];
    			char newc = sc.next().toCharArray()[0];
    			if (oldc == newc) continue;
    			d[index]=newc;
    			segs[(int)oldc-'a'].remove(index);
    			segs[(int)newc-'a'].add(index);
    			continue;
    		}
            int l = sc.nextInt()-1;
            int r = sc.nextInt()-1;
            int ans = 0;
            for (int j = 0; j < 26; j++) {
                if ((int)segs[j].ceiling(l) <= r) ans += 1;
            }
            System.out.println(ans);
    	}
    }
} 

