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: Accepted)

#include <bits/stdc++.h>
#define f first
#define s second
#define fore(i,a,b) for(int i = (a), ThxMK = (b); i < ThxMK; ++i)
#define pb push_back
#define all(s) begin(s), end(s)
#define _ ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define sz(s) int(s.size())
#define ENDL '\n'
using namespace std;
typedef long double ld;
typedef long long lli;
typedef pair<int,int> ii;
typedef vector<int> vi;
typedef vector<lli> vl;
typedef vector<ii> vii;
typedef vector<vi> vvi;
typedef vector<vl> vvl;
#define deb(x) cout << #x": " << (x) << endl;

const int N = 55;
const int INF = 1e9;
const lli MOD = 1e9+7;


vi primes(1,2);
vector<vector<ld>>nk;
void sieve(){
	vi m(N+1,0);
	for(int i = 4; i<N; i+=2)m[i]=1;
	for(int i = 3; i*i<=N; i+=2)if(!m[i])for(int j = i*i; j<N; j+=i)m[j]=1;
	for(int i = 3; i<N; i+=2)if(!m[i])primes.pb(i);
}

void pascal(){
	nk =vector<vector<ld>>(N,vector<ld>(N,0.0));
	fore(i,0,N)nk[i][0]=nk[i][i]=1.0;
	fore(i,1,N)fore(j,1,i)nk[i][j]=nk[i-1][j-1]+nk[i-1][j];
}

lli gcd(lli a, lli b){return (b?gcd(b,a%b):a);}

lli lcm(lli a, lli b){
	if(a<b)swap(a,b);
	lli c = gcd(a,b);
	a/=c;
	return a*b;
}

int popcount(lli x) { return __builtin_popcountll(x); }

lli poww(lli a, lli b){
	lli res =1;
	while(b){
		if(b&1) res = res * a;
		a = a*a;
		b/=2;
	}
	return res;
}

lli powm(lli a, lli b){
	lli res =1;
	while(b){
		if(b&1) res = (res * a)%MOD;
		a = (a*a)%MOD;
		b/=2;
	}
	return res;
}

bool isPrime(lli x){
	if(x==1)return 0;
	for(auto i:primes){
		if(i*i>x)return 1;
		if(x%i==0)return 0;
	}
	return 1;
}

lli SUM(lli a, lli b, lli mod=MOD){
	return (a+b+mod)%mod;
}
lli PROD(lli a, lli b, lli mod=MOD){
	a%=mod; b%=mod;
	return a*b % mod;
}
int dx[4]={-1,0,0,1}, dy[4]={0,1,-1,0};

// ---- コーディングはここから！ ('-')7



void solve(){
	string a,b; cin>>a>>b;
	int mn = sz(b);
	fore(i,0,sz(a)){
		int x = sz(b);
		fore(j,0,sz(b))if(i+sz(b)>sz(a))break;
		else x-=(a[i+j]==b[j]);
		mn = min(mn,x);
	}
	cout<<mn<<ENDL;

}

int main(){_
	//int t; cin>>t; while(t--)
	solve();
}