Dreamoon has a string s and a pattern string p. He first removes exactly x characters from s obtaining string s' as a result. Then he calculates <image> that is defined as the maximal number of non-overlapping substrings equal to p that can be found in s'. He wants to make this number as big as possible.

More formally, let's define <image> as maximum value of <image> over all s' that can be obtained by removing exactly x characters from s. Dreamoon wants to know <image> for all x from 0 to |s| where |s| denotes the length of string s.

Input

The first line of the input contains the string s (1 ≤ |s| ≤ 2 000).

The second line of the input contains the string p (1 ≤ |p| ≤ 500).

Both strings will only consist of lower case English letters.

Output

Print |s| + 1 space-separated integers in a single line representing the <image> for all x from 0 to |s|.

Examples

Input

aaaaa
aa


Output

2 2 1 1 0 0


Input

axbaxxb
ab


Output

0 1 1 2 1 1 0 0

Note

For the first sample, the corresponding optimal values of s' after removal 0 through |s| = 5 characters from s are {"aaaaa", "aaaa", "aaa", "aa", "a", ""}. 

For the second sample, possible corresponding optimal values of s' are {"axbaxxb", "abaxxb", "axbab", "abab", "aba", "ab", "a", ""}.
#include <bits/stdc++.h>
using namespace std;
clock_t start;
mt19937_64 rng(chrono::system_clock::now().time_since_epoch().count());
template <typename T>
void _F(const char *name, T arg1) {
  cerr << name << " = " << arg1 << '\n';
}
template <typename T, typename... Args>
void _F(const char *names, T arg1, Args... args) {
  const char *name = strchr(names, ',');
  cerr.write(names, name - names) << " = " << arg1 << '\n';
  _F(name + 2, args...);
}
template <typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &q) {
  in >> q.first >> q.second;
  return in;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &out, pair<T1, T2> &q) {
  out << q.first << " " << q.second;
  return out;
}
template <typename T1, typename T2>
pair<T1, T2> operator+(pair<T1, T2> p1, pair<T1, T2> p2) {
  return {p1.first + p2.first, p1.second + p2.second};
}
template <typename T1, typename T2>
pair<T1, T2> operator-(pair<T1, T2> p1, pair<T1, T2> p2) {
  return {p1.first - p2.first, p1.second - p2.second};
}
template <typename T1, typename T2>
bool operator<(pair<T1, T2> p1, pair<T1, T2> p2) {
  return p1 < p2;
}
template <typename T>
void Unique(vector<T> &v) {
  sort((v).begin(), (v).end()),
      v.resize(distance(v.begin(), unique((v).begin(), (v).end())));
}
const long long N = 2e3 + 1;
string s, p;
long long n, m;
long long dp[N][N], pre[N];
long long fun(long long i, long long skip) {
  if (i < skip || skip < 0) return -(long long)(1e9);
  if (i == skip) return 0;
  if (dp[i][skip] != -1) return dp[i][skip];
  dp[i][skip] = max(fun(i - 1, skip - 1), fun(i - 1, skip));
  if (pre[i] != -1) {
    dp[i][skip] = max(dp[i][skip], fun(pre[i], skip - (i - pre[i] - m)) + 1);
  }
  return dp[i][skip];
}
void solve(long long &test) {
  string s, p;
  cin >> s >> p;
  n = (long long)((s).size()), m = (long long)((p).size());
  memset(dp, -1, sizeof(dp));
  for (long long i = 1; i <= n; i++) {
    long long curs = i, curp = m;
    while (curs > 0 && curp > 0) {
      if (s[curs - 1] == p[curp - 1])
        curs--, curp--;
      else
        curs--;
    }
    if (curp == 0)
      pre[i] = curs;
    else
      pre[i] = -1;
  }
  for (long long i = 0; i <= n; i++) {
    cout << fun(n, i) << " ";
  }
  cout << '\n';
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  start = clock();
  cout << fixed << setprecision(20);
  for (long long i = 1; i <= t; ++i) {
    solve(i);
  }
  cerr << (double(clock() - start) / double(CLOCKS_PER_SEC));
  return 0;
}
