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
n=int(input())
S=list(input())
q=int(input())

D=[[0]*26 for _ in range(n+1)]
for i,s in enumerate(S):
  for j in range(26):
    if j==ord(s)-97:
      D[i+1][j]=D[i][j]+1
    else:
      D[i+1][j]=D[i][j]

n_max=n+1
nn=n_max.bit_length()+1
BITS=[[0]*(2**nn) for _ in range(26)]
for i in range(26):
  BITS[i].insert(0,0)

def bitsum(BIT,i):
  s=0
  while i>0:
    s+=BIT[i]
    i-=i&(-i)
  return s

def bitadd(BIT,i,x):
  while i<=2**nn:
    BIT[i]+=x
    i+=i&(-i)
  return BIT

for _ in range(q):
  q1,q2,q3=map(str,input().split())
  if q1=="1":
    num=int(q2)-1
    prev=ord(S[num])-97
    new=ord(q3)-97
    S[num]=q3
    if prev!=new:
      bitadd(BITS[prev],num+1,-1)
      bitadd(BITS[new],num+1,1)
  else:
    l,r=int(q2),int(q3)
    ans=0
    for i in range(26):
      diff=D[r][i]-D[l-1][i]+bitsum(BITS[i],r)-bitsum(BITS[i],l-1)
      if diff>0:
        ans+=1
    print(ans)

