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
class BIT:
  def __init__(self,n):
    self.s = [0]*(n+1)
    self.n = n
  
  def add(self,val,idx):
    while idx < self.n+1:
      self.s[idx] = self.s[idx] + val
      idx += idx&(-idx)
    return
  
  def get(self,idx):
    res = 0
    while idx > 0:
      res = res + self.s[idx]
      idx -= idx&(-idx)
    return res
 
import sys
sys.setrecursionlimit(10**6)
input = sys.stdin.readline
 
n = int(input())
S = [ord(s) - ord('a') for s in input()[:-1]]
 
bits = [BIT(n) for i in range(26)]
 
for i,s in enumerate(S):
  bits[s].add(1,i+1)
 
Q = int(input())
ans = []
for _ in range(Q):  
  flag,x,y = input().split()

  if flag == "1":
    i = int(x)
    c = ord(y) - ord('a')
    bits[S[i-1]].add(-1,i)
    bits[c].add(1,i)
    S[i-1] = c
  else:
    l = int(x)
    r = int(y)
    res = 0
    for i in range(26):
      res += (bits[i].get(r) > bits[i].get(l-1))
    ans.append(str(res))

print("\n".join(ans))