Problem:
Score : 200 points

Problem Statement

Given are two strings S and T.

Let us change some of the characters in S so that T will be a substring of S.

At least how many characters do we need to change?

Here, a substring is a consecutive subsequence. For example, xxx is a substring of yxxxy, but not a substring of xxyxx.

Constraints

The lengths of S and T are each at least 1 and at most 1000.

The length of T is at most that of S.

S and T consist of lowercase English letters.

Input

Input is given from Standard Input in the following format:

S
T

Output

Print the minimum number of characters in S that need to be changed.

Sample Input 1

cabacc
abc

Sample Output 1

1

For example, changing the fourth character a in S to c will match the second through fourth characters in S to T.

Since S itself does not have T as its substring, this number of changes - one - is the minimum needed.

Sample Input 2

codeforces
atcoder

Sample Output 2

6

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#define watch5(a,b,c,d,e) cerr<<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d << " | "<<#e<<": "<<e<<endl;
#define watch4(a,b,c,d) cerr<<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<endl;
#define watch3(a,b,c) cerr <<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<endl;
#define printclock cerr<<"Time : "<<1000*(ld)clock()/(ld)CLOCKS_PER_SEC<<"ms\n";
#define sharingan cerr << "\n-------------------------------------------\n\n";
#define watch2(a,b) cerr <<#a<<": "<<a<<" | "<<#b<<": "<<b<< endl;
#define Yup ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define watch(a) cerr <<#a<<": "<<a<<endl;
#define __gcd(a, b) __algo_gcd(a, b)
#define PI 3.14159265358979323846
#define mod 1000000007
#include "bits/stdc++.h"
using namespace std;
#define int long long

int32_t main(){
    string s, t;
    cin >> s >> t;
    int ans = INT_MAX;
    for( int i = 0; i < s.size(); i++ ){
        int cnt = 0;
        for( int j = 0, k = i; j < t.size(); j++, k++ ){
            if( t[j] != s[k] )cnt++;
        }
        ans = min(ans, cnt);
    }
    cout << ans;
}