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;
const int oo = 1000111000;
string s, p;
int f[2011][2011], res[2011];
void update(int &x, int val) { x = min(x, val); }
int main() {
  ios ::sync_with_stdio(false);
  cin.tie(NULL);
  cout << (fixed) << setprecision(6);
  while (cin >> s >> p) {
    int ls = s.length(), lp = p.length();
    string initial_p = p;
    while (p.length() < ls) p += initial_p;
    for (int i = (0), _b = (ls); i <= _b; i++)
      for (int j = (0), _b = (ls); j <= _b; j++) f[i][j] = oo;
    f[0][0] = 0;
    for (int i = (0), _b = (ls - 1); i <= _b; i++)
      for (int j = (0), _b = (i); j <= _b; j++)
        if (f[i][j] < oo) {
          if (s[i] == p[j]) {
            update(f[i + 1][j + 1], f[i][j]);
          }
          if (j % lp == 0) {
            update(f[i + 1][j], f[i][j]);
          }
          update(f[i + 1][j], f[i][j] + 1);
        }
    memset(res, 0, sizeof res);
    for (int i = (ls), _b = (ls); i <= _b; i++)
      for (int j = (0), _b = (i); j <= _b; j++) {
        if (j % lp == 0) {
          int nn = f[i][j];
          int ln = ls - j;
          for (int x = (nn), _b = (ln); x <= _b; x++)
            res[x] = max(res[x], j / lp);
        }
      }
    for (int i = (0), _b = (ls); i <= _b; i++) cout << res[i] << ' ';
    cout << endl;
  }
  return 0;
}
