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 SegT():
    def __init__(self, size):
        self.size = 1
        while self.size < size: self.size *= 2
        self.data = [0] * (self.size * 2 - 1)
    
    def update(self, i, x):
        i += (self.size - 1)
        self.data[i] = x
        while i > 0:
            i = (i - 1) // 2
            self.data[i] = self.data[i*2+1] | self.data[i*2+2]
        
    def get(self, a, b, node=0, l=0, r=-1):
        if r < 0: r = self.size
        if r <= a or b <= l: return 0
        if a <= l and r <= b: return self.data[node]
        vl = self.get(a, b, node*2+1, l, (l+r)//2)
        vr = self.get(a, b, node*2+2, (l+r)//2, r)
        return vl | vr

N = int(input())
S = input()
Q = int(input())

segt = SegT(N)
for i in range(N):
    x = ord(S[i]) - ord('a')
    segt.update(i, (1<<x))

for i in range(Q):
    t, l, r = input().split()
    if t == "1":
        x = ord(r) - ord('a')
        segt.update(int(l)-1, (1<<x))
    else:
        x = segt.get(int(l)-1,int(r))
        ans = 0
        for i in range(26):
            if x>>i&1: ans += 1
        print(ans)
