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;
inline int read() {
  int x;
  char c;
  while ((c = getchar()) < '0' || c > '9')
    ;
  for (x = c - '0'; (c = getchar()) >= '0' && c <= '9';) x = x * 10 + c - '0';
  return x;
}
char a[2000 + 5], b[500 + 5];
int p[2000 + 5], f[2000 + 5][2000 + 5];
inline void rw(int& x, int y) {
  if (y > x) x = y;
}
int main() {
  int n, m, i, j, k;
  scanf("%s%s", a + 1, b + 1);
  n = strlen(a + 1);
  m = strlen(b + 1);
  for (i = 1; i <= n; p[i++] = k)
    for (j = 1, k = i - 1; j <= m; ++j)
      for (++k; k <= n && a[k] != b[j];) ++k;
  memset(f, 200, sizeof(f));
  f[0][0] = 0;
  for (i = 0; i < n; ++i)
    for (j = 0; j <= n; ++j)
      if (f[i][j] >= 0)
        rw(f[i + 1][j], f[i][j]), rw(f[i + 1][j + 1], f[i][j]),
            p[i + 1] <= n ? rw(f[p[i + 1]][j + p[i + 1] - i - m], f[i][j] + 1),
            0             : 0;
  for (i = 0; i <= n; ++i) printf("%d ", f[n][i]);
}
