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>
char s[2001];
char t[501];
int bfr[2001];
int dp[2001][2001];
int ans[2001];
int min(int a, int b) { return a < b ? a : b; }
int main() {
  scanf(" %s", s);
  scanf(" %s", t);
  int n = strlen(s);
  int m = strlen(t);
  int i, j;
  for (i = 1; i <= n; i++) bfr[i] = -1;
  for (i = 0; i < n; i++) {
    int p = i;
    int q = m - 1;
    if (s[p] != t[q]) continue;
    while (p >= 0 && q >= 0) {
      while (p >= 0 && s[p] != t[q]) p--;
      if (p < 0) break;
      p--;
      q--;
    }
    if (q == -1) bfr[i + 1] = p + 1;
  }
  for (i = 0; i <= n; i++) dp[0][i] = 100000000;
  dp[0][0] = 0;
  for (i = 1; i <= n; i++) {
    for (j = 0; j <= n; j++) dp[i][j] = dp[i - 1][j];
    if (bfr[i] >= 0) {
      int res = i - bfr[i] - m;
      for (j = 1; j <= n; j++) {
        dp[i][j] = min(dp[i][j], dp[bfr[i]][j - 1] + res);
      }
    }
  }
  int pos = 0;
  while (pos <= n && dp[n][pos] != 100000000) pos++;
  pos--;
  int pre = 0;
  for (i = 1; i <= pos; i++) {
    for (j = pre; j < dp[n][i]; j++) ans[j] = i - 1;
    pre = dp[n][i];
  }
  int rem = n - pos * m;
  for (i = pre; i <= rem; i++) ans[i] = pos;
  pre = rem + 1;
  while (pos > 0) {
    pos--;
    for (i = 0; i < m; i++) ans[pre + i] = pos;
    pre += m;
  }
  for (i = 0; i <= n; i++) printf("%d ", ans[i]);
  puts("");
  return 0;
}
