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
from collections import defaultdict

class BIT(object):
    def __init__(self, N, init=0):
        self.N = N
        self.bit = [init for _ in range(N)]

    def add(self, k, w):
        x = k
        while x < self.N:
            self.bit[x] += w
            x |= x + 1

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

n = int(input())
s = list(input())
q = int(input())

d = defaultdict(lambda: BIT(n))
for i, si in enumerate(s):
    d[si].add(i, 1)

for i in range(q):
    k, x, y = input().split()
    if k == '1':
        x = int(x) - 1
        d[s[x]].add(x, -1)
        d[y].add(x, 1)
        s[x] = y
    else:
        l = int(x) - 1
        r = int(y) - 1
        ans = 0
        for v in d.values():
            ans += (v.sum(r + 1) - v.sum(l)) > 0
        print(ans)
