Make Them Equal
Theofanis has a string $$$s_1 s_2 \dots s_n$$$ and a character $$$c$$$. He wants to make all characters of the string equal to $$$c$$$ using the minimum number of operations.In one operation he can choose a number $$$x$$$ ($$$1 \le x \le n$$$) and for every position $$$i$$$, where $$$i$$$ is not divisible by $$$x$$$, replace $$$s_i$$$ with $$$c$$$. Find the minimum number of operations required to make all the characters equal to $$$c$$$ and the $$$x$$$-s that he should use in his operations.
Let's describe what happens in the third test case:   $$$x_1 = 2$$$: we choose all positions that are not divisible by $$$2$$$ and replace them, i. e. bzyx $$$\rightarrow$$$ bzbx;  $$$x_2 = 3$$$: we choose all positions that are not divisible by $$$3$$$ and replace them, i. e. bzbx $$$\rightarrow$$$ bbbb. 
The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.The first line of each test case contains the integer $$$n$$$ ($$$3 \le n \le 3 \cdot 10^5$$$) and a lowercase Latin letter $$$c$$$ — the length of the string $$$s$$$ and the character the resulting string should consist of.The second line of each test case contains a string $$$s$$$ of lowercase Latin letters — the initial string.It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.
For each test case, firstly print one integer $$$m$$$ — the minimum number of operations required to make all the characters equal to $$$c$$$.Next, print $$$m$$$ integers $$$x_1, x_2, \dots, x_m$$$ ($$$1 \le x_j \le n$$$) — the $$$x$$$-s that should be used in the order they are given.It can be proved that under given constraints, an answer always exists. If there are multiple answers, print any.
import sys
input = sys.stdin.readline

def solve():
    (n, c) = input().split()
    n = int(n)
    S = input().strip()
    if c * n == S:
        print(0)
        return
    for i in range(n // 2 + 1, n + 1):
        if S[i - 1] == c:
            print(1)
            print(i)
            return
    print(2)
    print(*[n - 1, n])
for _ in range(int(input())):
    solve()
import sys
input = sys.stdin.readline
for _ in range(int(input())):
    (n, c) = input().split(' ')
    n = int(n)
    c = c.strip()
    s = input().strip()
    if s == c * n:
        print(0)
    else:
        for index in range(n // 2, n):
            if s[index] == c:
                print(1)
                print(index + 1)
                break
        else:
            print(2)
            print(n, n - 1)
import heapq
from collections import Counter
import math
import sys
input = sys.stdin.readline

def inp():
    return int(input())

def inlt():
    return list(map(int, input().split()))

def insr():
    s = input()
    return list(s[:len(s) - 1])

def invr():
    return map(int, input().split())

def inis():
    return input().split()

def SieveOfEratosthenes(n):
    prime = [True for i in range(n + 1)]
    p = 2
    l = []
    while p * p <= n:
        if prime[p] == True:
            for i in range(p * p, n + 1, p):
                prime[i] = False
        p += 1
    for p in range(2, n + 1):
        if prime[p]:
            l.append(p)
    return l

def isPrime(n):
    prime_flag = 0
    if n > 1:
        for i in range(2, int(math.sqrt(n)) + 1):
            if n % i == 0:
                prime_flag = 1
                break
        if prime_flag == 0:
            return True
        else:
            return False
    else:
        return False
for _ in range(inp()):
    (p, c) = inis()
    n = int(p)
    s = insr()
    if s.count(c) == n:
        print(0)
        continue
    for i in range(n // 2, n):
        if s[i] == c:
            print(1)
            print(i + 1)
            break
    else:
        print(2)
        print(n, n - 1)