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
import bisect  # 配列二分法
from collections import defaultdict

N = int(input())
S = [*input()]
Q = int(input())
d = defaultdict(list)

# 各アルファベットの出現場所を記録
for i, c in enumerate(S):
    d[c] += [i]

# この時点でどのアルファベットについてもソート済み
# print(d)

for _ in range(Q):
    q, y, z = input().split()

    if q == '1':
        i = int(y) - 1  # 文字の変更位置

        if S[i] == z: continue  # 同じ文字の場合は処理終了

        # 削除 O(N)
        b = bisect.bisect(d[S[i]], i)  # 2分法で i の出現位置を(log N)で探す
        d[S[i]].pop(b - 1)  # 削除

        # 追加 O(N) かかる？
        S[i] = z
        bisect.insort(d[z], i) 

    else:
        left = int(y) - 1
        right = int(z) - 1

        count = 0
        for v in d.values():
            count += 1 if bisect.bisect(v, right) - bisect.bisect_left(v, left) > 0 else 0

        print(count)