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())
BIT=[[0]*n for i in range(26)]
w=[ord(x)-ord("a") for x in input()]
for i in range(n):
	k=i+1
	while k<=n:
		BIT[w[i]][k-1]+=1
		k=k+(k&-k)
q=int(input())
for i in range(q):
	a,b,c=input().split()
	if a=="1":
		k=int(b)
		while k<=n:
			BIT[ord(c)-ord("a")][k-1]+=1
			k=k+(k&-k)
		k=int(b)
		while k<=n:
			BIT[w[int(b)-1]][k-1]-=1
			k=k+(k&-k)	
		w[int(b)-1]=ord(c)-ord("a")
	else:
		b=int(b);c=int(c)
		ans=0
		for i in range(26):
			cnt=0
			k=c
			while k!=0:
				cnt+=BIT[i][k-1]
				k=k-(k&-k)
			k=b-1
			while k!=0:
				cnt-=BIT[i][k-1]
				k=k-(k&-k)
			if cnt>0:
				ans+=1
		print(ans)