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())
S = input()

L = 20
M = [1<<i for i in range(L)]
T = [[0]*m for m in M]

o = ord('a')

for i, c in enumerate(S):
    T[-1][i] = 1<<(ord(c)-o)

for l in range(L-2, -1, -1):
    for i in range(M[l]):
        T[l][i] = T[l+1][i*2] | T[l+1][i*2+1]
    
def q(layer, n, l, r):
    length_block = (1<<(L-1))//M[layer]
    ml = length_block*n
    mr = ml + length_block

    if l==ml and r==mr:
        return T[layer][n]
    
    center = ml + length_block//2

    ret = 0
    if l<center: ret |= q(layer+1, n*2  , l, min(r,center))
    if r>center: ret |= q(layer+1, n*2+1, max(l,center), r)

    return ret

Q = int(input())
for _ in range(Q):
    t, a1, a2 = input().split()
    if t=='1':
        i = int(a1)-1
        c = a2
        T[-1][i] = 1<<(ord(c)-o)
        for l in range(L-2, -1, -1):
            i = i//2
            T[l][i] = T[l+1][i*2] | T[l+1][i*2+1]
        
    else:
        l = int(a1)-1
        r = int(a2)
        x = q(0, 0, l, r)
        ans = 0
        for i in range(26):
            if (x>>i)&1: ans+=1
        print(ans)
        
