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
from bisect import bisect_left, bisect_right, insort_left

n = int(input())
s = list(input())
q = int(input())

d = {}
flag = {}
for i in list('abcdefghijklmnopqrstuvwxyz'):
	d.setdefault(i, []).append(-1)
	flag.setdefault(i, []).append(-1)
for i in range(n):
	d.setdefault(s[i], []).append(i)
for i in range(q):
	q1,q2,q3 = map(str,input().split())
	if q1 == '1':
		q2 = int(q2) - 1
		if s[q2] != q3:
			insort_left(flag[s[q2]],q2)
			insort_left(d[q3],q2)
			s[q2] = q3
	else:
		ans = 0
		q2 = int(q2) - 1
		q3 = int(q3) - 1
		if q2 == q3:
			print(1)
			continue
		for string,l in d.items():
			res = 0
			if d[string] != [-1]:
				left = bisect_left(l,q2)
				right = bisect_right(l,q3)
			else:
				left = 0
				right = 0
			if string in flag:
				left2 = bisect_left(flag[string],q2)
				right2 = bisect_right(flag[string],q3)
			else:
				left2 = 0
				right2 = 0
			if left != right:
				if right - left > right2 - left2:
					res = 1
			ans += res
			#print(string,l,res)
		print(ans)
