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();
        String s = sc.next();
        SegTree t = new SegTree(n, s);
        int q = sc.nextInt();
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < q; i++) {
            int c = sc.nextInt();
            if(c == 1){
                int a = sc.nextInt()-1;
                int b = sc.next().charAt(0) - 'a';
                t.update(a, b);
            }else{
                int l = sc.nextInt()-1;
                int r = sc.nextInt();
                int x = t.query(l, r, 0, 0, t.size);
                sb.append(Integer.bitCount(x) + "\n");
            }
        }
        System.out.print(sb.toString());
        sc.close();

    }

}

class SegTree {

    public int n;
    public int size;
    /**
     * 要素全体を含むノードのindex：0.
     * k番目(0-indexed)の要素のindex：k + size - 1
     * ノードkの親要素のindex：(k - 1) / 2.
     * ノードkの子要素のうち左側のindex：k*2+1.
     * ノードkの子要素のうち右側のindex：k*2+2.
     */
    public int[] data;

    /**
     * コンストラクタ.
     * @param n 要素数
     */
    public SegTree(int n, String s){
        this.n = n;
        size = 1;
        while(size < n) size *= 2;
        data = new int[size*2];
        // 初期化処理
        for (int i = 0; i < s.length(); i++) {
            int x = s.charAt(i) - 'a';
            update(i, x);
        }
    }

    /**
     * k番目（0-indexed）の要素をaに更新します.
     * @param k 変更対象要素のindex（0-indexed）
     * @param a 変更後の値
     */
    public void update(int k, int a){
        k += size - 1;
        if(data[k] != 1<<a){
            data[k] = 1<<a;
            while(0 < k){
                k = (k - 1) / 2;
                data[k] = data[k * 2 + 1] | data[k * 2 + 2];
            }
        }
    }

    /**
     * 区間[a, b)での○○を求めます.
     * 最初に呼ぶ際は、query(a, b, 0, 0, size)として呼ぶ.
     * @param a 求める区間の下限（含む）
     * @param b 求める区間の上限（含まない）
     * @param k 確認するノードのindex
     * @param l 確認するノードの下限（含む）
     * @param r 確認するノードの上限（含まない）
     * @return 区間[a, b)での○○
     */
    public int query(int a, int b, int k, int l, int r){
        if(r <= a || b <= l) return 0;

        if(a <= l && r <= b) return data[k];

        int vl = query(a, b, k*2+1, l, (l+r)/2);
        int vr = query(a, b, k*2+2, (l+r)/2, r);
        return vl | vr;
    }
}
