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.

-----Sample Input-----
7
abcdbbd
6
2 3 6
1 5 z
2 1 1
1 4 a
1 7 d
2 1 7

-----Sample Output-----
3
1
5

In the first query, cdbb contains three kinds of letters: b , c , and d, so we print 3.
In the second query, S is modified to abcdzbd.
In the third query, a contains one kind of letter: a, so we print 1.
In the fourth query, S is modified to abcazbd.
In the fifth query, S does not change and is still abcazbd.
In the sixth query, abcazbd contains five kinds of letters: a, b, c, d, and z, so we print 5.

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = "import bisect  # 配列二分法\nfrom collections import defaultdict\n\nN = int(input())\nS = [*input()]\nQ = int(input())\nd = defaultdict(list)\n\n# 各アルファベットの出現場所を記録\nfor i, c in enumerate(S):\n    d[c] += [i]\n\n# この時点でどのアルファベットについてもソート済み\n# print(d)\n\nfor _ in range(Q):\n    q, y, z = input().split()\n\n    if q == '1':\n        i = int(y) - 1  # 文字の変更位置\n\n        if S[i] == z: continue  # 同じ文字の場合は処理終了\n\n        # 削除 O(N)\n        b = bisect.bisect(d[S[i]], i)  # 2分法で i の出現位置を(log N)で探す\n        d[S[i]].pop(b - 1)  # 削除\n\n        # 追加 O(N) かかる？\n        S[i] = z\n        bisect.insort(d[z], i) \n\n    else:\n        left = int(y) - 1\n        right = int(z) - 1\n\n        count = 0\n        for v in d.values():\n            count += 1 if bisect.bisect(v, right) - bisect.bisect_left(v, left) > 0 else 0\n\n        print(count)"
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 14053