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 bisect

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)
#print(bag)

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