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
def main():
  import sys
  b=sys.stdin.buffer
  input=b.readline
  n=int(input())
  d=[0]*n+[1<<c-97for c in input()[:n]]
  for i in range(n-1,0,-1):d[i]=d[i+i]|d[i-~i]
  r=[]
  input()
  for q,a,b in zip(*[iter(b.read().split())]*3):
    i,s=int(a)+n-1,0
    if q<b'2':
      d[i]=1<<b[0]-97
      while i:
        i>>=1
        d[i]=d[i+i]|d[i-~i]
      continue
    j=int(b)+n
    while i<j:
      if i&1:
        s|=d[i]
        i+=1
      if j&1:
        j-=1
        s|=d[j]
      i>>=1
      j>>=1
    r+=bin(s).count('1'),
  print(' '.join(map(str,r)))
main()