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.io.PrintWriter;
import java.util.*;

public class Main {
    static Integer[] seg;
    static int size;
    static void update(int i, int x) {
        i += size;
        seg[i] = x;
        i >>= 1;
        while (i > 0) {
            seg[i] = seg[i * 2] | seg[i * 2 + 1];
            i >>= 1;
        }
    }
    static int query(int l, int r) {
        l += size;
        r += size;
        int s = 0;
        while (l < r) {
            if ((l & 1) == 1) {
                s |= seg[l];
                l++;
            }
            if ((r & 1) == 1) {
                r--;
                s |= seg[r];
            }
            l >>= 1;
            r >>= 1;
        }
        return Integer.bitCount(s);
    }
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        int n = Integer.parseInt(sc.next());
        size = n;
        seg = new Integer[size * 2];
        Arrays.fill(seg, 0);
        String s = sc.next();
        for (int i = 0; i < n; i++) update(i, 1 << s.charAt(i) - 'a');
        int q = Integer.parseInt(sc.next());
        for (int i = 0; i < q; i++) {
            int p = Integer.parseInt(sc.next());
            if (p == 1) {
                int j = Integer.parseInt(sc.next()) - 1;
                char c = sc.next().charAt(0);
                update(j, 1 << c - 'a');
            }
            else {
                int l = Integer.parseInt(sc.next()) - 1;
                int r = Integer.parseInt(sc.next());
                out.println(query(l, r));
            }
        }
        out.flush();
    }
}