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)

#include <iostream>
#include <math.h>
#include <numeric>
#include <vector>
#include <utility>
#include <algorithm>
#include <map>
#include <queue>
#include <queue>
#include <stack> 
#include <sstream>
#include <set>

typedef  long long ll;
const int dx[8]={1,0,-1,0,1,-1,-1,1};
const int dy[8]={0,1,0,-1,1,1,-1,-1};
const int dr[4] = {-1, 0, 1, 0};
const int dc[4] = { 0, 1, 0,-1};
const int INF = 1e9;
#define FOR(i, a, n) for(int i = (int)(a); i < (int)(n); ++i)
#define REP(i, n) FOR(i, 0, n)
#define SORT(a) sort(a.begin(),a.end())
#define REVERSE(a) reverse(a.begin(),a.end())
int guki (int a){
    if(a%2==0) return 0;
    else return 1;
}
int gcd(int a, int b){
    if(a%b==0){
        return b;
    }else{
        return (gcd(b,a%b));
    }
}
int lcm(int a, int b){
    int x = gcd(a,b);
    return (a*b/x);
}

using namespace std;
// now

struct UnionFind {
    vector<int> d;
    UnionFind(int n=0) : d(n,-1){}
    int Find (int x){//根の番号を探す
        if(d[x] < 0) return x;
        return d[x] = Find(d[x]);
    }
    bool unite (int x, int y){
        x = Find(x); y = Find(y);
        if(x == y ) return false;
        if(d[x] < d[y]){
            swap(x,y);
        }
        d[x] += d[y];
        d[y] = x;
        return true;
    } 

    bool same(int x, int y){
        return Find(x) == Find(y);
    }
    int size(int x) {
        return (-d[Find(x)]);
    }
};


ll nCr (ll n, ll r){
    ll ans = 1;
    for(ll i = n ; i > n-r ; i --){
        ans *= i;
    }
    for(ll i = 1 ; i<= r; i ++){
        ans /= i;
    }
    return ans ;
}

int nPr (int n,int r){
    int ans = 1;
    for(int i = n ; i > n-r ; i --){
        ans *= i;
    }
    return ans ;
}
 ll modd = 1e9+ 7;
ll wa (ll n,ll mi, ll ma){
    ll ans  = n * (mi+ma) /2;
    return ans;
}
ll kw(ll k,ll n){
    
    ll ans  =0;
    ll mi,ma;
    mi= wa(k,0,k-1);
    
    ma = wa(k,n-k+1,n);
    
    return (ma-mi +1)%modd;
}

int main(){
    string s,t;
    cin >> s >> t;

    int si = s.size(),ti = t.size();
    int mi = ti;

    for(int i=0;i<ti;i++){
        for(int j = i;j<ti;j++){
            string ss = t.substr(i,j-i+1);

            for(int k = 0;k < si-i-j+1;k++){
                string sss = s.substr(k,j-i+1);
                if(ss == sss){
                    if(i > k) continue;
                   if(si-(k+(j-i+1)) < ti -j-1) continue;
                    mi = min(mi,ti-(j-i+1));
                }
            }
        }
    }
    cout << mi <<endl;
}