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 MAX_L = 3000;
int T[MAX_L][MAX_L];
int last[MAX_L], z[MAX_L];
;
char s[MAX_L], p[MAX_L];
int main() {
  int Ls, Lp, is, ip, k, i, j, r;
  scanf("%s%s", &s, &p);
  Ls = strlen(s);
  Lp = strlen(p);
  for (i = 0; s[i]; i++) {
    is = i;
    ip = 0;
    while (s[is] && p[ip]) {
      while (s[is] && p[ip] && s[is] != p[ip]) is++;
      if (s[is] == p[ip]) {
        is++;
        ip++;
      }
    }
    if (!p[ip]) {
      last[i] = is - 1;
      z[i] = is - i - Lp;
    } else
      last[i] = -1;
  }
  if (last[0] != -1) T[last[0]][z[0]] = 1;
  for (i = 0; i < Ls - 1; i++)
    for (k = 0; k <= i + 1; k++) {
      j = i + 1;
      T[j][k] = max(T[j][k], T[i][k]);
      T[j][k + 1] = max(T[j][k + 1], T[i][k]);
      if (last[j] != -1) {
        r = last[j];
        T[r][k + z[j]] = max(T[r][k + z[j]], T[i][k] + 1);
      }
    }
  i = Ls - 1;
  for (j = 0; j <= Ls; j++) printf("%d ", T[i][j]);
  return 0;
}
