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
# -*- coding: utf-8 -*-

'''
7
abcdbbd
6
2 3 6
1 5 z
2 1 1
1 4 a
1 7 d
2 1 7
'''

# from helper import elapsed_time
from bisect import bisect_left, insort_left


N = int(input())
s = list(input())
Q = int(input())

# a-zまでの26文字に対して、s文字列の各文字に該当するindexを格納しておく
d = {i: [] for i in range(26)}
for i, c in enumerate(s):
    d[ord(c) - ord("a")].append(i)


# @elapsed_time
def convert(d: dict, s: list, q1: int, q2: str):
    if s[q1] == q2:
        return d, s

    didx_before = ord(s[q1]) - ord("a")
    didx_after = ord(q2) - ord("a")

    idx = bisect_left(d[didx_before], q1)
    d[didx_before].pop(idx)
    s[q1] = q2
    insort_left(d[didx_after], q1)

    return d, s


# @elapsed_time
def calc(d: dict, q1: int, q2: int):
    ans = 0
    for i in range(26):
        idx = bisect_left(d[i], q1)
        if d[i] and q1 <= d[i][-1] and d[i][idx] < q2 + 1:
            ans += 1

    return ans


for _ in range(Q):
    q, q1, q2 = input().split()
    q = int(q)
    if q == 1:
        d, s = convert(d, s, int(q1) - 1, q2)
    if q == 2:
        print(calc(d, int(q1) - 1, int(q2) - 1))
