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
sys.setrecursionlimit(10 ** 6)
input = sys.stdin.readline
N = int(input())
S = input().strip()
Q = int(input())
n = 1
while n < N:
    n *= 2
T = [0] * (2 * n)

def update(k, a):
    k += n-1
    T[k] = 1 << ord(a) - 97
    while k > 0:
        k = (k - 1) // 2
        T[k] = T[k*2+1] | T[k*2+2]

def query(a, b, k, l, r):
    # print(a, b, k, l, r)
    if r <= a or b <= l:
        # print('  ', set())
        return 0
    if a <= l and r <= b:
        # print('  ', T[k])
        return T[k]
    else:
        vl = query(a, b, k*2+1, l, (l+r)//2)
        vr = query(a, b, k*2+2, (l+r)//2, r)
        return vl | vr

for i in range(N):
    update(i, S[i])

for q in range(Q):
    f, i, c = [x for x in input().strip().split()]
    if f == '1':
        # print(T,T[int(i)-1])
        update(int(i)-1, c)
    elif f == '2':
        # print(int(i)-1, int(c)-1)
        tmp = query(int(i)-1, int(c), 0, 0, n)
        ans = 0
        for i in range(26):
            ans += tmp >> i & 1
         
        print(ans)
