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)

/*Author--Vishwas Agrawal*/
#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
#define vll vector<ll>
#define pb push_back
#define test ll t; cin>>t; while(t--)
#define fio ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define M 1000000007LL
#define all(x) x.begin(),x.end()
#define sortall(x) sort(all(x))
#define inc(i,a,n) for (ll i=a;i<n;i++)
#define dec(i,a,n) for (ll i=n-1;i>=a;i--)
long long binpow(long long base, long long exp, int mod) {
   long long res = 1;
   while (exp > 0) {
      if (exp % 2 == 1)
         res= (res * base) % mod;
      exp = exp >> 1;
      base = (base * base) % mod;
   }
   return res;
}
long long mod(long long x){
return ((x%M+M)%M);
}

long long add(long long a, long long b){
return mod(mod(a)+mod(b));
}

long long mul(long long a, long long b){
return mod(mod(a)*mod(b));
}

int main()
{
   string s,t;
   cin>>s>>t;
   ll min=1e5;
   inc(i,0,s.length()-t.length()+1)
   {
      ll temp=0;
      inc(j,0,t.length())
      {
         if(s[i+j]!=t[j])temp++;

      }
      if(temp<min)
      {
         min=temp;
      }
   }
   cout<<min<<endl;
   
  
   return 0;
}