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.InputStream;
import java.io.PrintStream;
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
import java.util.function.BinaryOperator;

class SegmentTree<T> {

	private List<T> nodes;
	private T e; // 単位元
	private int n;
	private BinaryOperator<T> operator;

	/**
	 * セグメント木を生成
	 *
	 * @param n  要素数
	 * @param e  単位元
	 * @param op
	 */
	public SegmentTree(int n, T e, BinaryOperator<T> op) {
		this.e = e;
		this.operator = op;

		n = 1 << (32 - Integer.numberOfLeadingZeros(Math.max(n - 1, 0))); // n以上で最小の2のべき乗
		this.n = n;

		nodes = new ArrayList<>(2 * n - 1);
		for (int i = 0; i < 2 * n - 1; i++) {
			nodes.add(e);
		}
	}

	//
	public void update(int k, T val) {
		k += n - 1;
		nodes.set(k, val);

		while (k > 0) {
			k = (k - 1) / 2;
			nodes.set(k, operator.apply(nodes.get(2 * k + 1), nodes.get(2 * k + 2)));
		}
	}

	// 区間 [a,b) のoperatorの結果を返す
	// 0-indexed
	public T query(int a, int b) {
		return query(a, b, 0, 0, n);
	}

	private T query(int a, int b, int k, int l, int r) {
		if (a >= r || b <= l) {
			return e;
		}

		if (a <= l && r <= b) {
			return nodes.get(k);
		}

		T vl = query(a, b, 2 * k + 1, l, (l + r) / 2);
		T vr = query(a, b, 2 * k + 2, (l + r) / 2, r);
		return operator.apply(vl, vr);
	}

}

public class Main {
	InputStream in = System.in;
	PrintStream out = System.out;

	static private List<Character> merge(List<Character> a, List<Character> b) {
		List<Character> r = new ArrayList<>();

		int ai = 0;
		int bi = 0;

		for (;;) {
			if (ai >= a.size()) {
				for (; bi < b.size(); bi++) {
					r.add(b.get(bi));
				}
				break;
			}
			if (bi >= b.size()) {
				for (; ai < a.size(); ai++) {
					r.add(a.get(ai));
				}
				break;
			}

			if (a.get(ai) == b.get(bi)) {
				r.add(a.get(ai));
				ai++;
				bi++;
			} else if (a.get(ai) < b.get(bi)) {
				r.add(a.get(ai));
				ai++;
			} else {
				r.add(b.get(bi));
				bi++;
			}

		}

		return r;
	}

	public void _main(String[] args) {
		Scanner sc = new Scanner(in);

		int N = sc.nextInt();
		String S = sc.next();
		int Q = sc.nextInt();

		SegmentTree<Integer> seg = new SegmentTree<>(N, 0, (a, b) -> (a|b));

		for (int i = 0; i < S.length(); i++) {
			
			int c = S.charAt(i) - 'a';
			int b = 1<<c;
			
			seg.update(i, b);
		}

		for (int q = 0; q < Q; q++) {
			int t = sc.nextInt();

			if (t == 1) {
				int i = sc.nextInt() - 1;
				char c = sc.next().charAt(0);

				int b = 1 << (c - 'a');
				

				seg.update(i, b);

			} else {
				int l = sc.nextInt() - 1;
				int r = sc.nextInt();

				int ans = seg.query(l, r);
				out.println(Integer.bitCount(ans));
			}
		}

		sc.close();

	}

	public static void main(String[] args) {
		new Main()._main(args);
	}
}
