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.Scanner;
import java.util.TreeSet;

public class Main {
  public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);

    int N = scanner.nextInt();
    char[] S = scanner.next().toCharArray();
    TreeSet<Integer>[] trees = new TreeSet[26];
    for (int i = 0; i < 26; i++) trees[i] = new TreeSet<>();
    for (int i = 0; i < N; i++) trees[S[i] - 'a'].add(i);

    int Q = scanner.nextInt();
    for (int i = 0; i < Q; i++) {
      int type = scanner.nextInt();
      switch (type) {
        case 1:
          int p = scanner.nextInt() - 1;
          char c = scanner.next().charAt(0);
          trees[S[p] - 'a'].remove(p);
          trees[c - 'a'].add(p);
          S[p] = c;
          break;
        case 2:
          int l = scanner.nextInt() - 1;
          int r = scanner.nextInt() - 1;
          int cnt = 0;
          for (int j = 0; j < 26; j++) {
            Integer left = trees[j].ceiling(l);
            if (left != null && left <= r) cnt++;
          }
          System.out.println(cnt);
          break;
      }
    }
  }
}
