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 sys
input = sys.stdin.readline


class BIT:
    def __init__(self, size):
        self.bit = [0]*(size+1)
        self.size = size

    def sum(self, i):
        i += 1
        s = 0
        while i > 0:
            s += self.bit[i]
            i -= i & -i
        return s

    def add(self, i, x):
        i += 1
        while i <= self.size:
            self.bit[i] += x
            i += i & -i


N = int(input())
S = list(input().rstrip())
bits = [BIT(N) for _ in range(26)]
for i in range(N):
    bits[ord(S[i])-ord('a')].add(i, 1)
Q = int(input())
for _ in range(Q):
    t, x, y = input().split()
    if t == '1':
        i = int(x) - 1
        c = y
        bits[ord(S[i])-ord('a')].add(i, -1)
        bits[ord(c)-ord('a')].add(i, 1)
        S[i] = c
    else:
        l = int(x) - 1
        r = int(y) - 1
        ans = 0
        for i in range(26):
            cnt = bits[i].sum(r) - (bits[i].sum(l-1) if l > 0 else 0)
            if cnt > 0:
                ans += 1
        print(ans)
