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
# -*- coding: utf-8 -*-
import sys

N=input()
S=["None"]+list(raw_input())
Q=input()
query=[ sys.stdin.readline().split() for _ in range(Q) ]

#BIT
bit=[ [ 0 for _ in range(N+1) ] for _ in range(27) ]  #1-indexed bit[1]:aのbit, bit[2]:bのbit ...

def add(idx,a,w):
    while a<=N:
        bit[idx][a]+=w
        a+=a&-a

def sum(idx,a):
    ret=0
    while 0<a:
        ret+=bit[idx][a]
        a-=a&-a
    return ret

for i,x in enumerate(S):
    if i==0: continue
    add(ord(x)-96,i,1)  #a~zのどのbit配列か,特定文字のbit配列の何番目,bitに代入する値


for q1,q2,q3 in query:
    q1=int(q1)
    q2=int(q2)
    if q1==1:
        before_q3=S[q2] #文字列Sの変更前の文字をbefore_q3に代入
        S[q2]=q3        #文字列Sの変更部分1文字を更新
        q3=ord(q3)-96                   #変更後の文字をアスキーコード10進数で表現
        before_q3=ord(before_q3)-96     #変更前の文字をアスキーコード10進数で表現
        add(q3,q2,1)                    #a~zのどのbit配列か,特定文字のbit配列の何番目,bitに代入する値
        add(before_q3,q2,-1)    #置き代えた文字は-1しなきゃいけない
    elif q1==2:
        q3=int(q3)
        cnt=0
        for i in range(1,27):
            if 0<sum(i,q3)-sum(i,q2-1):
                cnt+=1
        print cnt