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, n):
        self.size = n
        self.tree = [0] * (n + 1)
 
    def sum(self, i):
        if i == 0:
            return 0
        s = 0
        while i > 0:
            s += self.tree[i]
            i -= i & -i
        return s
 
    def add(self, i, x):
        while i <= self.size:
            self.tree[i] += x
            i += i & -i
aas = "abcdefghijklmnopqrstuvwxyz"
N = int(input().strip())
S = [" "]+list(input().strip())
Q = int(input().strip())
cbt = dict()
for c in aas:
    cbt[c] = Bit(N)
    for i, a in enumerate(S):
        if i == 0:
            continue
        if a == c:
            cbt[c].add(i, 1)

for _ in range(Q):
    t, i, c = input().split()
    t = int(t)
    if t == 1:
        i = int(i)
        if S[i] != c:
            cbt[S[i]].add(i, -1)
            cbt[c].add(i, 1)
            S[i] = c
    else:
        l, r = int(i), int(c)
        s = 0
        for a in aas:
            bt = cbt[a]
            s += (bt.sum(r)-bt.sum(l-1)>0)
        print(s)
