B. Crossword solving（Round422）

B. Crossword solving

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Erelong Leha was bored by calculating of the greatest common divisor of two factorials. Therefore he decided to solve some crosswords. It's well known that it is a very interesting occupation though it can be very difficult from time to time. In the course of solving one of the crosswords, Leha had to solve a simple task. You are able to do it too, aren't you?

Leha has two strings s and t. The hacker wants to change the string s at such way, that it can be found in t as a substring. All the changes should be the following: Leha chooses one position in the string s and replaces the symbol in this position with the question mark "?". The hacker is sure that the question mark in comparison can play the role of an arbitrary symbol. For example, if he gets strings="ab?b" as a result, it will appear in t="aabrbb" as a substring.

Guaranteed that the length of the string s doesn't exceed the length of the string t. Help the hacker to replace in s as few symbols as possible so that the result of the replacements can be found in t as a substring. The symbol "?" should be considered equal to any other symbol.

Input

The first line contains two integers n and m (1 ≤ n ≤ m ≤ 1000) — the length of the string s and the length of the string tcorrespondingly.

The second line contains n lowercase English letters — string s.

The third line contains m lowercase English letters — string t.

Output

In the first line print single integer k — the minimal number of symbols that need to be replaced.

In the second line print k distinct integers denoting the positions of symbols in the string s which need to be replaced. Print the positions in any order. If there are several solutions print any of them. The numbering of the positions begins from one.

Examples

input

3 5
abc
xaybz

output

2
2 3 

input

4 10
abcd
ebceabazcd

output

1
2 

 hint：问你最少换几个字符，可以让s变成t的子串

大力搞一发

直接从t的头开始往后匹配判断找最小就好了。。。

    #include<iostream>
    #include<cstdio>
    #include<stdlib.h>
    #include<string>
    #include<string.h>
    #include<vector>
    #include<stack>
    #include<queue>
    #include<map>
    #include<cmath>
    using namespace std;
    #define pi acos(-1.0)
    typedef long long ll;

    int key[1050];
    int main(){
        memset(key, -1, sizeof(key));
        int n, m;
        cin >> n >> m;
        string s, t;
        cin >> s >> t;
       // cout << s << " " << t << endl;
       // int len1 = s.size(), len2 = t.size();
        if(n == m){
            int cnt = 0;
            for(int i = 0; i < n; i++)
                if(s[i] != t[i]){
                    key[cnt] = i + 1;
                    cnt++;
            }
            cout << cnt << endl;
            for(int i=0; i<cnt; i++){
                cout << key[i] << " ";
            }
            cout << endl;
        }else{
            int Min = 1050, cnt = 0;
            for(int i=0; i<=m-n; i++){
                cnt = 0;
                for(int j=i; j<i+n; j++){
                    if(s[j-i] != t[j]){
                        cnt++;
                    }
                }
                if(cnt < Min){
                    Min = cnt;
                    int ind = 0;
                    for(int j=i; j<i+n; j++){
                        if(s[j-i] != t[j]){
                            key[ind++] = j - i + 1;
                        }
                    }
                }
            }
            cout << Min << endl;
            for(int i=0; i<Min; i++){
                cout << key[i] << " ";
            }
            cout << endl;
        }

        return 0;
    }