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;
void open() {
  freopen("string.in", "r", stdin);
  freopen("string.out", "w", stdout);
}
class party {
 public:
  char s[3000], p[1000];
  int A[3000], f[3000][3000], ls, lp;
  void work() {
    scanf("%s%s", s + 1, p + 1);
    ls = strlen(s + 1);
    lp = strlen(p + 1);
    pre_process();
    top_process();
    for (int i = 0; i < ls; ++i) printf("%d ", f[ls][i]);
    printf("%d\n", f[ls][ls]);
  }

 private:
  void pre_process() {
    int t;
    for (int pos = 1; pos <= ls; ++pos) {
      t = pos - 1;
      for (int i = 1; i <= lp; ++i) {
        ++t;
        while (s[t] != p[i] && t <= ls) ++t;
      }
      A[pos] = t;
    }
  }
  void top_process() {
    for (int i = 1; i <= ls; ++i) {
      for (int j = 0; j <= min(ls, i - 1); ++j) {
        f[i + 1][j] = max(f[i + 1][j], f[i][j]);
        f[i + 1][j + 1] = max(f[i + 1][j + 1], f[i][j]);
        f[A[i]][j + A[i] - i + 1 - lp] =
            max(f[A[i]][j + A[i] - i + 1 - lp], f[i - 1][j] + 1);
      }
    }
  }
} diary;
int main() {
  diary.work();
  return 0;
}
