/*
You are given two strings, a main string S, and a pattern P. You have to find the starting indices of the anagrams of P in S.


Anagrams are permutations of a string. For P="abc”, its anagrams are abc,acb,bca,bac,cba,cab.


Note that indexing starts at 0.

Input format
There is one line of input, containing two space-separated strings S and P.

Output format
First-line should contain the number of such starting indices.

Next line should contain the indices in increasing order separated by a space.

Sample Input 1
aaba ab

Sample Output 1
2

1 2

Explanation 1
The anagrams of pattern "ab" are “ab” and “ba”. These are present at indices 1 and 2 of the input string “aaba”.

Sample Input 2
bacdgabcda abcd

Sample Output 2
3

0 5 6

Explanation 2
The anagrams of "abcd" can be seen as “bacd” at index 0, “abcd” at index 5 and “bcda” at index 6.

Constraints
1 <= length(S), length(P) <= 10^6

*/


/**
 * @param {string} s
 * @param {string} p
 * @return {number[]}
 */
// sliding window of frequency array: https://www.youtube.com/watch?v=MW4lJ8Y0xXk&t=202s
function findAllAnagramsInAString(s, p) {
  let hash= new Array(26).fill(0);
  let currentWindow=new Array(26).fill(0);
  let ans=[];
  if(p.length>s.length)
  return ans;
  // console.log(getIndex("e"));
  for(let i=0;i<p.length;i++){
    hash[getIndex(p[i])]++;
  }
  // console.log(hash);
  // get first window
  for(let i=0;i<p.length;i++){
    let currentIndex=getIndex(s[i]);
    currentWindow[currentIndex]++;
  }
  if(compare(hash,currentWindow))
    ans.push(0);
  for(let i=p.length;i<s.length;i++){
    let currentIndex=getIndex(s[i]);
    let removedCharacterIndex=getIndex(s[i-p.length]);
    currentWindow[currentIndex]++;
    currentWindow[removedCharacterIndex]--;
    if(compare(hash,currentWindow))
      ans.push(i-p.length+1);
  }


  return ans;

}

function main() {
  let [s, p] = readLine().split(" ");
  let result = findAllAnagramsInAString(s, p);

  console.log(result.length);

  console.log(...result);
}