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)

//#pragma GCC optimize("Ofast")
//#pragma GCC target("avx,avx2,fma")
#pragma GCC optimize("O3")
#include<iostream>
#include<algorithm>
#include<string>
#include<cmath>
#include<cstdlib>
#include<set>
#include<vector>
#include<sstream>
#include<queue>
#include <iomanip>
#include<unordered_set>
#include <stack>
#include <unordered_map>
#include <map>
#include<numeric>
#include<chrono>
#include <fstream>
constexpr auto INF = 9223372036854775807;
typedef long long int ll;
typedef unsigned long long int ull;
typedef unsigned long int ul;
#define f(i,a,b) for(ll i=(ll)a;i<(ll)b;i+=1)
#define rf(i,a,b) for(ll i=(ll)a;i>=(ll)b;i-=1)
#define endl '\n'
#define N 1000000007 // prime modulo value 
#define M 998244353
#define all(x) x.begin(),x.end()
#define mkp(a,b) make_pair(a,b)
using namespace std;
struct custom_hash {
	static uint64_t splitmix64(uint64_t x) {
		// http://xorshift.di.unimi.it/splitmix64.c
		x += 0x9e3779b97f4a7c15;
		x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
		x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
		return x ^ (x >> 31);
	}

	size_t operator()(uint64_t x) const {
		static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
		return splitmix64(x + FIXED_RANDOM);
	}
};
inline int mex(unordered_set<int> st)
{
	int m = 0;
	while (st.find(m) != st.end())
	{
		m += 1;
	}
	return m;
}
inline int calculateGrundy(int n)
{
	if (n == 1)
		return 0;
	unordered_set<int> mexi;
	for (int i = 1; i <= sqrt(n); i += 1)
	{
		if (n % i == 0)
		{
			if (n / i != n)
				mexi.insert(calculateGrundy(n / i));
			mexi.insert(calculateGrundy(i));
		}
	}
	return mex(mexi);
}
inline ll gcd(ll a, ll b)
{
	// This is code taken from geeksforgeek
	if (b == 0)
		return a;
	return gcd(b, a % b);
}
/*ll power(ll x, ll y)
{
	// This is code taken from geeksforgeek
	ll res = 1;

	x = x % N;
	if (x == 0) return 0;
	while (y > 0)
	{
		if (y & 1)
			res = (res * x) % N;
		y = y >> 1;
		x = (x * x) % N;
	}
	return res;
}*/
ll power(ll x, ll y)
{
	ll res = 1;

	x = x;

	while (y > 0)
	{

		if (y & 1)
			res = (res * x);

		y = y >> 1;
		x = (x * x);
	}
	return res;
}


/*ll modInverse(ll n, ll p)
{
	return power(n, p - 2, p);
}*/

ll sum(ll n)
{
	return (n * (n + 1)) / 2;
}
ll sum_digit(ll n)
{
	ll ans = 0;
	while (n)
	{
		ans += n % 10;
		n /= 10;
	}
	return ans;
}
ll gp(ll r, ll n)
{
	return (pow(r,n)-1)/(r-1);
}
int main()
{
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	string s,t;
	cin>>s>>t;
	ll ans=INF;
	f(i,0,s.size()-t.size()+1)
	{
		ll temp=0;
		f(j,0,t.size())
		{
			if(s[i+j]!=t[j])
			{
				temp+=1;
			}
		}
		ans=min(ans,temp);
	}
	cout<<ans;
	return 0;
}