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;
string S, P;
int LS, LP;
int dpdp[2002][2002][2];
int res[2002];
void solve() {
  int i, j, k, l, r, x, y;
  string s;
  cin >> S >> P;
  LS = S.size();
  LP = P.size();
  memset(dpdp, 0xff, sizeof(dpdp));
  for (i = 0; i < LS + 1; i++)
    for (j = 0; j < LS + 1; j++) dpdp[i][j][0] = 3000;
  for (i = 0; i < LS + 1; i++)
    for (j = 0; j < LS + 1; j++) dpdp[i][j][1] = -1;
  dpdp[0][0][0] = dpdp[0][0][1] = 0;
  for (i = 0; i < LS; i++)
    for (j = 0; j < LS; j++)
      if (dpdp[i][j][1] >= 0) {
        if (S[i] == P[j % LP]) {
          dpdp[i + 1][j + 1][0] = min(dpdp[i + 1][j + 1][0], dpdp[i][j][0]);
          dpdp[i + 1][j + 1][1] = max(dpdp[i + 1][j + 1][1], dpdp[i][j][1]);
        }
        if ((j % LP) == 0)
          dpdp[i + 1][j][0] = min(dpdp[i + 1][j][0], dpdp[i][j][0]);
        dpdp[i + 1][j][0] = min(dpdp[i + 1][j][0], dpdp[i][j][0] + 1);
        dpdp[i + 1][j][1] = max(dpdp[i + 1][j][1], dpdp[i][j][1] + 1);
      }
  for (i = 0; i < LS + 1; i++)
    if (dpdp[LS][i][1] >= 0) {
      for (j = dpdp[LS][i][0]; j <= dpdp[LS][i][1]; j++)
        res[j] = max(res[j], i / LP);
    }
  for (i = 0; i < LS + 1; i++) (void)printf("%d ", res[i]);
  (void)printf("\n");
}
int main(int argc, char** argv) {
  string s;
  int i;
  if (argc == 1) ios::sync_with_stdio(false);
  for (i = 0; i < argc - 1; i++) s += argv[i + 1], s += '\n';
  for (i = 0; i < s.size(); i++) ungetc(s[s.size() - 1 - i], stdin);
  solve();
  return 0;
}
