You are given a string s consisting of lowercase Latin letters and q queries for this string.

Recall that the substring s[l; r] of the string s is the string s_l s_{l + 1} ... s_r. For example, the substrings of "codeforces" are "code", "force", "f", "for", but not "coder" and "top".

There are two types of queries: 

  * 1~ pos~ c (1 ≤ pos ≤ |s|, c is lowercase Latin letter): replace s_{pos} with c (set s_{pos} := c); 
  * 2~ l~ r (1 ≤ l ≤ r ≤ |s|): calculate the number of distinct characters in the substring s[l; r]. 

Input

The first line of the input contains one string s consisting of no more than 10^5 lowercase Latin letters.

The second line of the input contains one integer q (1 ≤ q ≤ 10^5) — the number of queries.

The next q lines contain queries, one per line. Each query is given in the format described in the problem statement. It is guaranteed that there is at least one query of the second type.

Output

For each query of the second type print the answer for it — the number of distinct characters in the required substring in this query.

Examples

Input


abacaba
5
2 1 4
1 4 b
1 5 b
2 4 6
2 1 7


Output


3
1
2


Input


dfcbbcfeeedbaea
15
1 6 e
1 4 b
2 6 14
1 7 b
1 12 c
2 6 8
2 1 6
1 7 c
1 2 f
1 10 a
2 7 9
1 10 a
1 14 b
1 1 f
2 1 11


Output


5
2
5
2
6
import os, sys, math
import collections


#res = solve('(R' + ('(R)R' * 2) + ')')

if os.path.exists('testing'):
	name = os.path.basename(__file__)
	if name.endswith('.py'):
		name = name[:-3]
	src = open(name + '.in.txt', encoding='utf8')
	input = src.readline

tree_size_bits = int(math.floor(math.log2(10 ** 5))) + 1
letter_count = ord('z') - ord('a') + 1
def solve(s, commands):
	tree = [ 0 ] * (2 ** (tree_size_bits + 1))
	data = [ None ] * len(s)

	def set_char(pos, c):
		full_mask = (1 << letter_count) - 1
		remove_mask = (1 << data[pos]) if data[pos] is not None else 0
		add_index = ord(c) - ord('a')
		add_mask = 1 << add_index

		if remove_mask == add_mask:
			return
		data[pos] = add_index
		
		pos = 2 ** tree_size_bits + pos
		tree[pos] = (tree[pos] & (full_mask ^ remove_mask)) | add_mask
		while pos > 1:
			parent = pos // 2
			child1 = parent * 2 + 0
			child2 = parent * 2 + 1

			new_value = tree[child1] | tree[child2]
			if tree[parent] == new_value:
				break
			tree[parent] = new_value
			pos = parent
	def query(pos1, pos2):
		def query_rec(array_index, lower_bound, upper_bound):
			if pos1 <= lower_bound and upper_bound <= pos2:
				return tree[array_index]
			if pos1 > upper_bound or pos2 < lower_bound:
				return 0
			split_value = (lower_bound + upper_bound + 1) // 2
			return query_rec(array_index * 2 + 0, lower_bound, split_value - 1) | query_rec(array_index * 2 + 1, split_value, upper_bound)
		return query_rec(1, 0, 2 ** tree_size_bits - 1)

	for e, char in enumerate(s):
		set_char(e, char)

	result = []
	for c, a, b in commands:
		if c == '1':
			pos = int(a) - 1
			char = b
			set_char(pos, char)
		else:
			lower_bound = int(a) - 1
			upper_bound = int(b) - 1
			characters = query(lower_bound, upper_bound)
			count = 0
			while characters:
				if characters & 1:
					count += 1
				characters = characters >> 1
			result.append(count)
	return result

s = input().strip()
num = int(input().strip())
commands = [ input().strip().split() for _ in range(num) ]
r = solve(s, commands)
for q in r:
	print(q)

#abacaba
#5
#2 1 4
#1 4 b
#1 5 b
#2 4 6
#2 1 7

# 3
