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())
s = input()
table = {x: 2**i for i, x in enumerate(map(chr, range(97, 123)))}

SEG_LEN = n
SEG = [0]*(SEG_LEN*2)


def update(i, x):
    i += SEG_LEN
    SEG[i] = table[x]
    while i > 0:
        i //= 2
        SEG[i] = SEG[i*2] | SEG[i*2+1]


def find(left, right):
    left += SEG_LEN
    right += SEG_LEN
    ans = 0
    while left < right:
        if left % 2 == 1:
            ans = SEG[left] | ans
            left += 1
        left //= 2
        if right % 2 == 1:
            ans = SEG[right-1] | ans
            right -= 1
        right //= 2
    return format(ans, 'b').count('1')


for i, c in enumerate(s):
    update(i, c)

q = int(input())
for _ in range(q):
    com, *query = input().split()
    if com == '1':
        idx, x = query
        update(int(idx)-1, x)
    else:
        L, R = map(int, query)
        print(find(L-1, R))
