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
# -*- coding: utf-8 -*-

#############
# Libraries #
#############

import sys
input = sys.stdin.readline

N = int(input())
S = list(input())
Q = int(input())


alphabet = "abcdefghijklmnopqrstuvwxyz"

bag = { l:[] for l in alphabet}

for i in range(N):
  bag[S[i]].append(i+1)


import bisect

for _ in range(Q):
  t,a,b = input().split()
  a = int(a)
  if int(t) == 1:
    if S[a-1]!=b:
      bag[S[a-1]].remove(a)
      S[a-1] = b
      bisect.insort(bag[b],a)
  else:
    count = 0
    b = int(b)
    for l in alphabet:
      if len(bag[l])>0:
        left = bisect.bisect_left(bag[l],a)
        if left != len(bag[l]):
          if bag[l][left] <= b:
            count += 1
    print(count)
