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

TO_ACII = 97

N = int(input())
S = ["0"] + list(input())
Q = int(input())

ABC = [[]for i in range(0,26,1)]#該当文字a:0 b:1 ...z:26 が何文字目に出てきたかを保持する配列

for i in range(1,N+1,1):
    ABC[ord(S[i])-TO_ACII].append(i)

ans = []
for i in range(0,Q,1):
    q = list(input().split())
    if q[0]=="1":#文字変更
        changed = S[int(q[1])]
        ABC[ord(changed)-TO_ACII].pop(bisect_left(ABC[ord(changed)-TO_ACII],int(q[1])))
        ABC[ord(q[2])-TO_ACII].insert(bisect_left(ABC[ord(q[2])-TO_ACII],int(q[1])),int(q[1]))
        S[int(q[1])]=q[2]
    else:
        tmpans = 0
        for i in range(0,26,1):

            if bisect_right(ABC[i],int(q[1])-1)<len(ABC[i]) and int(q[2]) >= ABC[i][bisect_right(ABC[i],int(q[1])-1)]:
                tmpans+=1
        ans.append(tmpans)

for i in range(0,len(ans),1):
    print(ans[i])
