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 "bits/stdc++.h"
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#ifdef on_linux
	#include <sys/resource.h>
#endif

#ifdef dobby_is_a_free_elf
	#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
#else
	#define trace(...) 0;
#endif

using namespace std;
using namespace __gnu_pbds;


#define all(x)                  (x).begin(), (x).end()
#define between(i,x,y)          ((i) >= (x) && (i) <= (y))
#define DECIMAL(n)              cout << fixed << setprecision(n)
#define fi                      first
#define lbd                     lower_bound
#define mem0(x)                 memset(x, 0, sizeof (x))
#define mem1(x)                 memset(x, -1, sizeof (x))
#define mp                      make_pair
#define pb                      push_back
#define pf                      push_front
#define ppb                     pop_back
#define ppf                     pop_front
#define sc                      second
#define sqr(a)                  ((a) * (a))
#define sz(x)                   (int)((x).size())
#define ubd                     upper_bound

#define int                     long long

using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;

template <typename T>
using ordered_set = tree<T, null_type,
      less<T>,
      rb_tree_tag,
      tree_order_statistics_node_update>;
// find_by_order(k)  returns iterator to kth element starting from 0
// order_of_key(k) returns count of elements strictly smaller than k


const long double PI = 3.14159265358979323846264338;
const double eps = 1e-10;
#ifdef int
	const long long INF = 1e18;
#else
	const long long INF = 1e9;
#endif
const int32_t fftmod = 998244353;
const int32_t MOD = 1000000007;
const int32_t N = 1e5 + 10;
const int32_t NN = 1e6 + 10;



inline void elf_magic(int testcase)
{
	string s, t;
	cin >> s >> t;
	int ans = sz(t);
	for (int i = 0; i < sz(s); ++i) {
		int x = 0;
		for (int j = 0; j < sz(t); ++j) {
			x += (s[i + j] != t[j]);
		}
		ans = min(ans, x);
	}
	cout << ans;
	return;
}

int32_t main()
{
	#ifdef on_linux
	rlimit cpu_time {.rlim_cur = 2, .rlim_max = RLIM_INFINITY};
	setrlimit(RLIMIT_CPU, &cpu_time);
	#endif

	ios_base::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.tie(nullptr);

	int TESTS = 1;
	//cin >> TESTS;
	for (int i = 1; i <= TESTS; ++i) {
		// cout << "Case #" << i << ": ";
		elf_magic(i);
		if (i != TESTS)	{ cout << '\n'; }
	}
	return 0;
}