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())

#アルファベットの各文字に対してからのリストを持つ辞書
alpha={chr(i):[] for i in range(97,123)}

#alpha[c]で各文字ごとに出現場所をソートして保管
for i,c in enumerate(s):
    bisect.insort(alpha[c],i+1)

for i in range(int(input())):
    p,q,r=input().split()
    if p=='1':
        q=int(q)
        b=s[q-1]
        if b!=r:
            alpha[b].pop(bisect.bisect_left(alpha[b],q))
            bisect.insort(alpha[r],q)
            s[q-1]=r
    else:
        count=0
        for l in alpha.values():
            pos=bisect.bisect_left(l,int(q))
            if pos<len(l) and l[pos]<=int(r):
                count+=1
        print(count)
