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
'''input
abacaba
5
2 1 4
1 4 b
1 5 b
2 4 6
2 1 7
'''
from sys import stdin
from collections import deque
import math



# builds the segment tree.... time complexity O(n)
def build_segment_tree(tree, string, index, start, end):
	if start > end:
		return

	# for node element
	if start == end:
		tree[index][ord(string[start]) - ord('a')] += 1

	# for internal node
	else:
		mid = (start + end) // 2
		build_segment_tree(tree, string, 2 * index, start, mid)
		build_segment_tree(tree, string, 2 * index + 1, mid + 1, end)
		for i in range(26):
			tree[index][i] = tree[2 * index][i] + tree[2 * index + 1][i]


def combine(first, second):
	aux = [0] * 26
	for i in range(26):
		aux[i] = first[i] + second[i]
	return aux


# query function. Outputs the minimum element in the given range. Time complexity is O(log(n))
def query(tree, a, index, start, end, qs, qe):
	if qs > end or qe < start:
		return [0] * 26

	# total overlap
	if qs <= start and end <= qe:
		return tree[index][:]
	# partial overlap
	else:
		mid = (start + end) // 2
		left = query(tree, a, 2 * index, start, mid, qs, qe)
		right = query(tree, a, 2 * index + 1, mid + 1, end, qs, qe)
		return combine(left, right)



# update the given node. Time complexity is O(log(n))
def update_node(tree, string, index, start, end, n_index, c):
	# No overlap
	if n_index < start or n_index > end:
		return

	# reach the node
	if start == end:
		tree[index][ord(string[n_index]) - ord('a')] -= 1
		tree[index][ord(c) - ord('a')] += 1
		string[n_index] = c
		return

	# partial or complete overlap
	else:
		mid = (start + end) // 2
		update_node(tree, string, 2 * index, start, mid, n_index, c)
		update_node(tree, string, 2 * index + 1, mid + 1, end, n_index, c)
		aux = combine(tree[ 2 * index], tree[ 2 * index + 1])
		for i in range(26):
			tree[index][i] = aux[i]
		return


# main starts
string = list(stdin.readline().strip())
n = len(string)
tree = [[0 for x in range(26)] for y in range(4 * n + 1)]

index = 1
start = 0
end = n - 1
build_segment_tree(tree, string, index, start, end)
# print(tree[1])
q = int(stdin.readline().strip())
for _ in range(q):
	t, x, y = list(stdin.readline().split())

	if t == '1':
		update_node(tree, string, index, start, end, int(x) - 1, y)
		# print(tree[1])
	else:
		result = query(tree, string, index, start, end, int(x) - 1, int(y) - 1)
		# print(result)
		count = 0
		for i in result:
			if i > 0:
				count += 1
		print(count)