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>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse4")
using namespace std;
constexpr long long mod = 1000000007;
const long long INF = mod * mod;
const double eps = 1e-12;
const double pi = acosl(-1.0);
long long mod_pow(long long x, long long n, long long m = mod) {
  long long res = 1;
  while (n) {
    if (n & 1) res = res * x % m;
    x = x * x % m;
    n >>= 1;
  }
  return res;
}
struct modint {
  long long n;
  modint() : n(0) { ; }
  modint(long long m) : n(m) {
    if (n >= mod)
      n %= mod;
    else if (n < 0)
      n = (n % mod + mod) % mod;
  }
  operator int() { return n; }
};
bool operator==(modint a, modint b) { return a.n == b.n; }
modint operator+=(modint& a, modint b) {
  a.n += b.n;
  if (a.n >= mod) a.n -= mod;
  return a;
}
modint operator-=(modint& a, modint b) {
  a.n -= b.n;
  if (a.n < 0) a.n += mod;
  return a;
}
modint operator*=(modint& a, modint b) {
  a.n = ((long long)a.n * b.n) % mod;
  return a;
}
modint operator+(modint a, modint b) { return a += b; }
modint operator-(modint a, modint b) { return a -= b; }
modint operator*(modint a, modint b) { return a *= b; }
modint operator^(modint a, long long n) {
  if (n == 0) return modint(1);
  modint res = (a * a) ^ (n / 2);
  if (n % 2) res = res * a;
  return res;
}
long long inv(long long a, long long p) {
  return (a == 1 ? 1 : (1 - p * inv(p % a, a)) / a + p);
}
modint operator/(modint a, modint b) { return a * modint(inv(b, mod)); }
const int max_n = 1 << 10;
modint fact[max_n], factinv[max_n];
void init_f() {
  fact[0] = modint(1);
  for (int i = 0; i < max_n - 1; i++) {
    fact[i + 1] = fact[i] * modint(i + 1);
  }
  factinv[max_n - 1] = modint(1) / fact[max_n - 1];
  for (int i = max_n - 2; i >= 0; i--) {
    factinv[i] = factinv[i + 1] * modint(i + 1);
  }
}
modint comb(int a, int b) {
  if (a < 0 || b < 0 || a < b) return 0;
  return fact[a] * factinv[b] * factinv[a - b];
}
void solve() {
  string s, p;
  cin >> s >> p;
  int n = s.size();
  int m = p.size();
  int d = n / m;
  vector<vector<int>> dp(d + 1, vector<int>(m));
  vector<vector<int>> cop(d + 1, vector<int>(m));
  for (int i = 0; i < d + 1; i++)
    for (int j = 0; j < m; j++) dp[i][j] = mod;
  cop = dp;
  dp[0][0] = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < d + 1; j++)
      for (int k = 0; k < m; k++) {
        if (dp[j][k] == mod) continue;
        if (s[i] == p[k]) {
          int nj = j, nk = k + 1;
          if (nk == m) {
            nj++;
            nk = 0;
          }
          cop[nj][nk] = min(cop[nj][nk], dp[j][k]);
        }
        int nj = j, nk = 0;
        cop[nj][nk] = min(cop[nj][nk], dp[j][k]);
        cop[j][k] = min(cop[j][k], dp[j][k] + 1);
      }
    for (int j = 0; j < d + 1; j++)
      for (int k = 0; k < m; k++) {
        dp[j][k] = cop[j][k];
        cop[j][k] = mod;
      }
  }
  vector<int> ans(n + 1);
  for (int j = 0; j < d + 1; j++)
    for (int k = 0; k < m; k++) {
      if (dp[j][k] <= n) {
        ans[dp[j][k]] = max(ans[dp[j][k]], j);
      }
    }
  for (int i = 1; i <= n; i++) {
    ans[i] = max(ans[i], ans[i - 1]);
  }
  for (int i = 0; i < n + 1; i++) {
    int r = n - i;
    ans[i] = min(ans[i], r / m);
    if (i > 0) cout << " ";
    cout << ans[i];
  }
  cout << "\n";
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  solve();
  return 0;
}
