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

N = int(input())
S = list(str(input()))

def ordN(x):
  return ord(x) - ord('a')

li = [[] for _ in range(26)]
for i,s in enumerate(S):
  li[ordN(s)].append(i)

for i in range(int(input())):
  Q1, Q2, Q3 = input().split()
  if Q1 == "1":
    Q2 = int(Q2) - 1
    if S[Q2] != Q3:
      I = bisect.bisect_left(li[ordN(S[Q2])], Q2)
      li[ordN(S[Q2])].pop(I)
      bisect.insort(li[ordN(Q3)], Q2)
      S[Q2] = Q3
  else:
    Q2, Q3 = int(Q2)-1, int(Q3)-1
    count = 0
    for j in range(26):
      I = bisect.bisect_left(li[j], Q2)
      if I < len(li[j]) and li[j][I] <= Q3:
        count += 1
    print(count)