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
#Range Sum Query
n=int(input())
s=input()
q=int(input())
#1番から始まる数え方
#要素数　Seg_Len
Seg_Len=1<<19
seg=[0 for i in range(Seg_Len*2)]
#i番目の要素はi+Seg_Len-1
def add(ind,v):
    ind+=Seg_Len-1 #元々の配列のインデックスとセグメント木の長さを足すとセグメント木の配列内のインデックスになる。
    seg[ind]=v  #一番下の部分に値を追加する。
    while(1):
        ind//=2
        if(ind==0):
            break  #a[0]にアクセスさせない
        seg[ind]=seg[ind*2]|seg[ind*2+1]
        
def sum1(l,r):
    l+=Seg_Len-1
    r+=Seg_Len-1
    ans=0
    while(l<r):
        if(l%2==1):
            ans|=seg[l]
            l+=1
        l//=2
        if(r%2==1):
            ans|=seg[r-1]
            r-=1
        r//=2
    tmp=0
    while(ans):
        if(ans & 1):
            tmp+=1
        ans>>=1
    return tmp
#abc と　124の対応付け
alp='abcdefghijklmnopqrstuvwxyz'
d={}
for i in range(len(alp)):
    d[alp[i]]=2**i
    
for i in range(len(s)):
    add((i+1),d[s[i]])
for i in range(q):
    com,a,b=input().split()
    if(com=='1'):
        ind=int(a)
        v=d[b]
        add(ind,v)
    if(com=='2'):
        l=int(a)
        r=int(b)
        ##rは一つ増やしておく必要がある
        print(sum1(l,r+1))
        