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.n = n
        self.bit = [0]*(n+1)
    
    def add(self, i, x):
        i += 1
        
        while i<=self.n:
            self.bit[i] += x
            i += i&(-i)

    def acc(self, i):
        s = 0
        
        while i>0:
            s += self.bit[i]
            i -= i&(-i)
        
        return s

N = int(input())
S = list(input()[:-1])
bits = [BIT(N) for _ in range(26)]

for i in range(N):
    j = ord(S[i])-ord('a')
    bits[j].add(i, 1)

Q = int(input())

for _ in range(Q):
    q = input().split()
    
    if q[0]=='1':
        i, c = int(q[1])-1, q[2]
        j = ord(S[i])-ord('a')
        bits[j].add(i, -1)
        k = ord(c)-ord('a')
        bits[k].add(i, 1)
        S[i] = c
    else:
        l, r = int(q[1])-1, int(q[2])-1
        ans = 0
        
        for i in range(26):
            if bits[i].acc(r+1)-bits[i].acc(l)>0:
                ans += 1
        
        print(ans)