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 sys
input = sys.stdin.readline

N=int(input())
S=list(input().strip())
Q=int(input())

LEN=N

BIT=[[0]*(LEN+1) for i in range(26)]

def update(v,w,i): # index vにwを加える
    while v<=LEN:
        BIT[i][v]+=w
        v+=(v&(-v)) # 自分を含む大きなノードへ. たとえばv=3→v=4

def getvalue(v,i): # [1,v]の区間の和を求める
    ANS=0
    while v!=0:
        ANS+=BIT[i][v]
        v-=(v&(-v)) # 自分より小さい2ベキのノードへ. たとえばv=3→v=2へ
    return ANS

for i in range(N):
    update(i+1,1,ord(S[i])-97)

for i in range(Q):
    qf,x,y=input().split()
    qf=int(qf)

    if int(qf)==1:
        x=int(x)
        update(x,-1,ord(S[x-1])-97)
        update(x,1,ord(y)-97)
        S[x-1]=y

    else:
        x=int(x)
        y=int(y)
        ANS=0
        for i in range(26):
            if getvalue(y,i)-getvalue(x-1,i)>0:
                ANS+=1

        print(ANS)
        
    
