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
def popcount(x):
    return bin(x).count('1')

def initialize(N):
    i=1
    while i<N:
        i*=2
    return [0]*(2*i-1),i

def update(i,x):
    i+=d-1
    bit=ord(x)-97
    SEG[i]=0 | (1<<bit)
    while i>0:
        i=(i-1)//2
        SEG[i]=SEG[i*2+1]|SEG[i*2+2]
        
def find(a,b,k,l,r):
    if r<=a or b<=l:
        return 0
    if a<=l and r<=b:
        return SEG[k]
    else:
        c1=find(a,b,2*k+1,l,(l+r)//2)
        c2=find(a,b,2*k+2,(l+r)//2,r)
        return c1|c2
        
N=int(input())
S=input()
Q=int(input())
SEG,d=initialize(N)
for i in range(N):
    update(i,S[i])
for i in range(Q):
    com,s,t=map(str,input().split())
    if(com=='1'):
        update(int(s)-1,t)
    else:
        print(popcount(find(int(s)-1,int(t),0,0,d)))