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;
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  string S, P;
  cin >> S >> P;
  int N = S.length(), M = P.length();
  vector<int> nxt(N + 1, N + M + 47);
  for (int i = 0; i < N; i++) {
    int a = 0;
    for (int j = i; j < N; j++) {
      if (P[a] == S[j]) a++;
      if (a == M) {
        nxt[i] = j + 1;
        break;
      }
    }
  }
  vector<vector<int> > A(N + 1, vector<int>(N + 1, N + 2));
  A[N][0] = 0;
  for (int i = N - 1; i >= 0; i--) {
    for (int j = 0; j <= N; j++) A[i][j] = A[i + 1][j];
    for (int j = 0; j < N; j++)
      if (nxt[i] <= N)
        A[i][j + 1] = min(A[i][j + 1], nxt[i] - i - M + A[nxt[i]][j]);
  }
  vector<int> ans(N + 1, 0);
  for (int j = 0; j <= N; j++)
    if (A[0][j] <= N) ans[A[0][j]] = max(ans[A[0][j]], j);
  for (int i = 0; i < N; i++) {
    if (N - i - ans[i] * M > 0)
      ans[i + 1] = max(ans[i + 1], ans[i]);
    else
      ans[i + 1] = max(ans[i + 1], ans[i] - 1);
  }
  for (int i = 0; i <= N; i++) cout << ans[i] << ((i == N) ? "\n" : " ");
  return 0;
}
