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
N=int(input())
from functools import reduce
size=int(N**.5+(1-1e-9))
a=[0]*N
b=[0]*size
f=lambda v: reduce(lambda x,y:x|y,v,0)
def init(v):
    a=v
    b=[f(a[j*size:(j+1)*size]) for j in range(size)]
    return a,b
def update(i,v):
    a[i]=v
    j=i//size
    b[j]=f(a[j*size:(j+1)*size])
def r_query(l,r):
    L,R=0--l//size,r//size
    if L>=R:
        return f(a[l:r])
    else:
        return f(a[l:L*size]+b[L:R]+a[R*size:r])
def c(n):
    r=0
    while n:
        r+=n&1
        n>>=1
    return r

o=lambda c:ord(c)-ord('a')
bit=[2**i for i in range(26)]
S=input()
Q=int(input())
XYZ=[input().split() for i in range(Q)]
a,b=init([bit[o(c)] for c in S])
for x,y,z in XYZ:
    if x=='1':
        update(int(y)-1,bit[o(z)])
    else:
        print(c(r_query(int(y)-1,int(z))))