Codeforces Round #422 (Div. 2) B. Crossword solving 枚举-优快云博客

本文介绍了一个字符串匹配问题，通过枚举方法找到最少需要替换的字符数，使得一个字符串能作为子串出现在另一个字符串中。该方法适用于编程竞赛中的字符串处理题目。

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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

题意：

　　给你两个字符串S,T

　　问你至少将S串中几个字符改成问号，使得能在T串中找到S串

　　问号可以和任何字符匹配

题解：

　　枚举i,j 表示在S串中 i 位置起点， T串中j 位置为起点是匹配完成的位置

　　将必须是问号的pos，数量，记录下来，取个最小的

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
const long long INF = 1e18+1LL;
const double pi = acos(-1.0);
const int N = 1e3+5, M = 1e3+20,inf = 2e9;


int n,m;
char s[N],t[N];
vector<int > G[N];
int main() {
    scanf("%d%d",&n,&m);
    scanf("%s%s",s+1,t+1);
    int mi = inf;
    for(int i = 1; i <= m - n + 1; ++i) {
        for(int j = 1; j <= n; ++j) {
            if(s[j] != t[i+j-1]) {
                G[i].push_back(j);
            }
        }
        mi = min(mi,(int)G[i].size());
    }
    for(int i = 1; i <= m - n + 1; ++i) {
        if(G[i].size() == mi) {
            cout<<mi<<endl;
            for(int j = 0; j < G[i].size(); ++j) {
                cout<<G[i][j]<<" ";
            }
            return 0;
        }
    }
    return 0;
}