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
#A1 ~ Aiまでの和 O(logN)
def BIT_query(BIT,idx):
    res_sum = 0
    while idx > 0:
        res_sum += BIT[idx]
        idx -= idx&(-idx)
    return res_sum

#Ai += x O(logN)
def BIT_update(BIT,idx,x):
    while idx <= n:
        BIT[idx] += x
        idx += idx&(-idx)
    return


alps = 'abcdefghijklmnopqrstuvwxyz'

n = int(input())
s = list(input())
char_hd_d = {}
for alp in alps:
    char_hd_d[alp] = [0]*(n+1)

for i, si in enumerate(s):
    BIT_update(char_hd_d[si], i+1, 1)

q = int(input())
ansl = []
for _ in range(q):
    command, a, b = input().split()
    if command == '1':
        i = int(a)
        c = b
        i_char = s[i-1]
        BIT_update(char_hd_d[i_char], i, -1)
        BIT_update(char_hd_d[c], i, 1)
        s[i-1] = c 
    else:
        l,r = int(a),int(b)
        cnt = 0
        for alp in alps:
            if BIT_query(char_hd_d[alp], r) - BIT_query(char_hd_d[alp], l-1) > 0:
                cnt += 1
        ansl.append(cnt)

for a in ansl:
    print(a)            