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
def update(i,x):
    i += d-1
    bit = ord(x)-97
    seg[i] = 0 | (1<<bit)
    while i > 0:
        i = (i-1)//2
        seg[i] = seg[i*2+1]|seg[i*2+2]
def find(a,b,k,l,r):
    if r <= a or b <= l:
        return 0
    if a <= l and r <= b:
        return seg[k]
    c1 = find(a,b,2*k+1,l,(l+r)//2)
    c2 = find(a,b,2*k+2,(l+r)//2,r)
    return c1|c2
n = int(input())
s = input()
q = int(input())
d = 1
while d < n:
    d *= 2
seg = [0]*(2*d-1)
for i in range(n):
    update(i,s[i])
for i in range(q):
    type,a,b = input().split()
    if type == "1":
        update(int(a)-1,b)
    else:
        part = find(int(a)-1,int(b),0,0,d)
        print(bin(part).count("1"))
