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())

def ctoi(c):
  return ord(c)-97
class Bit:
  def __init__(self, n):
    self.size = n
    self.tree = [[0]*(n+1) for _ in range(26)]
  def query(self,a,i):
    r = 0
    while i > 0:
      r += self.tree[a][i]
      i -= i & -i
    return r
  def update(self,i,a,x):
    while i<=n:
      self.tree[a][i]+=x
      i+=i&(-i)
bi=Bit(n)
for i in range(n):
  bi.update(i+1,ctoi(s[i]),1)
  
for _ in range(q):
  a,b,c=input().split()
  a,b=int(a),int(b)
  if a==1 and s[b-1]!=c:
    bi.update(b,ctoi(s[b-1]),-1)
    bi.update(b,ctoi(c),1)
    s[b-1]=c
  elif a==2:
    c=int(c)
    right=[bi.query(i,c) for i in range(26)]
    left=[bi.query(i,b-1) for i in range(26)]
    ans=0
    for r,l in zip(right,left):
      if r-l>0:
        ans+=1
    print(ans)