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())[: -1]
Q = int(input())

class BIT:
  def __init__(self, n):
    self.n = n
    self.data = [0] * (n + 1)
    self.el = [0] * (n + 1)
  def sum(self, i):
    s = 0
    while i > 0:
      s += self.data[i]
      i -= i & -i
    return s
  def add(self, i, x):
    self.el[i] += x
    while i <= self.n:
      self.data[i] += x
      i += i & -i
  def get(self, i, j = None):
    if j is None:
      return self.el[i]
    return self.sum(j) - self.sum(i)

fwks = [BIT(N + 1) for _ in range(26)]
a = ord("a")
for i in range(N):
  j = ord(S[i]) - a
  #print(j)
  fwks[j].add(i + 1, 1)
for _ in range(Q):
  q, x, y = input().split()
  q = int(q)
  if q == 1:
    x = int(x)
    for i in range(26):
      if fwks[i].get(x): fwks[i].add(x, -1)
    fwks[ord(y) - a].add(x, 1)
  else:
    x = int(x)
    y = int(y)
    res = 0
    #print(x, y)
    for i in range(26):
      res += fwks[i].get(x - 1, y) > 0
    print(res)