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
import sys
def input(): return sys.stdin.readline().rstrip()

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

char_idx = [[] for _ in range(26)]
for i in range(n):
    char_idx[ord(s[i])-ord('a')].append(i)
# print(char_idx)

query = [input().split() for _ in range(q)]
# print(query)

for t, a, b in query:
    if t=='1':
        i = int(a) - 1
        if s[i] == b:
            continue
        else:
            idx = bisect.bisect_left(char_idx[ord(s[i])-ord('a')], i)
            if char_idx[ord(s[i])-ord('a')][idx] == i:
                del char_idx[ord(s[i])-ord('a')][idx]
            bisect.insort_left(char_idx[ord(b)-ord('a')], i)
            s[i] = b
    else:
        l = int(a) - 1
        r = int(b) - 1
        ans = 0
        for i in range(26):
            if len(char_idx[i]) == 0:
                continue
            it = bisect.bisect_left(char_idx[i], l)
            if it < len(char_idx[i]) and char_idx[i][it] <= r:
                ans += 1
        print(ans)
