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 = [x for x in input()]
Q = int(input())

#A1 ... AnのBIT(1-indexed)
BIT = [[0] * (N + 1) for _ in range(26)]

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

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

for i, c in enumerate(S):
    BIT_update(BIT[ord(c)-ord('a')], i+1, 1)
  
for _ in range(Q):
    a, b, c = input().rstrip().split()
    if int(a) == 1:
        BIT_update(BIT[ord(S[int(b)-1])-ord('a')], int(b), -1)
        BIT_update(BIT[ord(c)-ord('a')], int(b), 1)
        S[int(b)-1] = c
    else:
        count = 0
        for i in range(26):
            if BIT_query(BIT[i], int(b)-1) != BIT_query(BIT[i], int(c)):
                count += 1
        print(count)