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;
char s[2010], p[2010];
int dp[2010][2010], lst[2010], pr[2010], ans[2010];
int main() {
  scanf("%s%s", s, p);
  int n = strlen(s), m = strlen(p);
  memset(pr, -1, sizeof(pr));
  for (int i = 0; i < n; i++) {
    if (s[i] != p[m - 1]) {
      lst[i] = -1;
      continue;
    }
    int ii = i;
    for (int j = m - 1; j > -1; j--) {
      for (; ii >= 0 && s[ii] != p[j]; ii--)
        ;
      if (ii < 0) {
        lst[i] = -1;
        break;
      }
      ii--;
    }
    if (lst[i] > -1) lst[i] = ii + 1;
  }
  for (int i = 0; i < n; i++) {
    if (lst[i] < 0) continue;
    for (int j = 0; j < lst[i]; j++)
      if (lst[j] > -1) pr[i] = j;
  }
  int t0, t1;
  memset(dp, -1, sizeof(dp));
  for (int i = 0; i < n; i++) dp[i][0] = 0;
  for (int i = 0; i < n; i++) {
    if (i)
      for (int j = 0; j < n + 1; j++) dp[i][j] = dp[i - 1][j];
    if (lst[i] < 0) continue;
    t0 = i - lst[i] + 1 - m;
    if (lst[i] > 0)
      for (int j = 1; j < n + 1; j++) {
        t1 = dp[lst[i] - 1][j - 1];
        if (t1 > -1 && (dp[i][j] == -1 || dp[i][j] > t1 + t0))
          dp[i][j] = t1 + t0;
      }
    else
      dp[i][1] = t0;
  }
  for (int i = 0; i < n + 1; i++) {
    t0 = dp[n - 1][i];
    if (t0 < 0) continue;
    for (int j = t0; j < n - m * i + 1; j++) ans[j] = max(ans[j], i);
  }
  for (int i = 0; i < n + 1; i++) printf("%d%c", ans[i], i < n ? ' ' : '\n');
  return 0;
}
