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
n=int(input())
ss=input()
dps=['a']*n
from collections import defaultdict
d=defaultdict(list)
for i in range(n):
    d[ss[i]].append(i)
    dps[i]=ss[i]
sorted(d.items())

Q=int(input())
import bisect
for i in range(Q):
    num,ii,c=input().split()
    #print(d)
    if num=='1':
        if c==dps[int(ii)-1]:
            continue
        bisect.insort_left(d[c],int(ii)-1)
        d[dps[int(ii)-1]].remove(int(ii)-1)
        dps[int(ii)-1]=c
    else:
        l,r=int(ii)-1,int(c)-1
        count=0
        for k,lst in d.items():
            idx=bisect.bisect_left(lst,l)
            idx1=bisect.bisect_right(lst,r)
            count+=((idx1-idx)>0)
            #print(lst,l,r,count)
        print(count)
        pass