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
#!/usr/bin/python3
import bisect

n = int(input())
s = input()
q = int(input())

idxs = [ [] for i in range(26) ]
sl = [ s[i] for i in range(n) ]

for i in range(n):
    idxs[ord(s[i]) - ord('a')].append(i + 1)

#print(idxs)

qs = []
for i in range(q):
    qs.append(input())

for qi in qs:
    (op, x, y) = qi.split()
    if op == '1':
        j = int(x)
        c = y
        oc = sl[j - 1]
        if oc != c:
            sl[j - 1] = c
            idx = bisect.bisect_left(idxs[ord(oc) - ord('a')], j)
            del idxs[ord(oc) - ord('a')][idx]
            idx = bisect.bisect_left(idxs[ord(c) - ord('a')], j)
            idxs[ord(c) - ord('a')].insert(idx, j)
        
    elif op == '2':
        l = int(x)
        r = int(y)
        c = 0
        for i in range(26):
            il = bisect.bisect_left(idxs[i], l)
            ir = bisect.bisect_right(idxs[i], r)
            if il < ir:
                c += 1
        print(c)
