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)

#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;
typedef long long int lld;

const lld N = 200043;
const lld MOD = 1000000007;

lld add(lld x, lld y)
{
    x =((x%MOD)+(y%MOD))%MOD;
    while(x >= MOD) x -= MOD;
    while(x < 0) x += MOD;
    return x;
}

lld mul(lld x, lld y)
{
    return ((x%MOD)*(y%MOD))% MOD;
}

lld binpow(lld x, lld y)
{
    lld z = 1;
    while(y)
    {
        if(y & 1) z = mul(z, x);
        x = mul(x, x);
        y >>= 1;
    }
    return z;
}

lld inv(lld x)
{
    return binpow(x, MOD - 2);
}

lld divide(lld x, lld y)
{
    return mul(x, inv(y));
}

// Combinations

/*
lld fact[N];

void precalc()
{
    fact[0] = 1;
    for(lld i = 1; i < N; i++)
        fact[i] = mul(fact[i - 1], i);
}

lld C(lld n, lld k)
{
    return divide(fact[n], mul(fact[k], fact[n - k]));
}
*/
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    lld t,i,j;
    //cin>>t;
    t=1;
    while(t--)
    {
        string str,ptr;
        cin>>str>>ptr;
        lld ans=LLONG_MAX;
        for(lld i=0;i<str.length();i++)
        {

            lld ctr=0;
            if(i+ptr.length()>=str.length())
                break;
            for(lld j=i;j<i+ptr.length();j++)
                if(str[j]!=ptr[j-i])
                ctr++;
            ans=min(ans,ctr);
            //cout<<ctr<<endl;
        }
        cout<<ans<<endl;
    }

}