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 l[2010];
int nxt[2010][30], dp[2010][2010];
int n, m;
int cal(int x) {
  for (int i = 0; i < m; i++) {
    x = nxt[x][p[i] - 'a'];
    if (x == -1) return 0x3f3f3f3f;
    x++;
  }
  return x - 1;
}
void Max(int &x, int y) { x = max(x, y); }
int main() {
  scanf("%s%s", s + 1, p);
  n = strlen(s + 1);
  m = strlen(p);
  memset(nxt, -1, sizeof(nxt));
  for (int i = n; i >= 1; i--) {
    for (int j = 0; j < 26; j++) nxt[i][j] = nxt[i + 1][j];
    nxt[i][s[i] - 'a'] = i;
  }
  for (int i = 1; i <= n; i++) l[i] = cal(i);
  memset(dp, 0, sizeof(dp));
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= i; j++) {
      if (i < n && l[i + 1] <= n)
        Max(dp[l[i + 1]][j + l[i + 1] - i - m], dp[i][j] + 1);
      if (i < n) Max(dp[i + 1][j + 1], dp[i][j]), Max(dp[i + 1][j], dp[i][j]);
    }
  }
  for (int i = 0; i <= n; i++) {
    printf("%d", dp[n][i]);
    if (i == n)
      puts("");
    else
      printf(" ");
  }
}
