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 N = 2e3 + 7, inf = 0x3f3f3f3f;
int dp[N][N], n, m, i, j, k, len, p, cnt, x;
char s[N], ss[N];
int cal(int x) {
  for (p = m, cnt = 0, k = x; k && p; --k)
    if (ss[p] == s[k])
      p--;
    else
      cnt++;
  if (!p)
    return cnt;
  else
    return inf;
}
int main() {
  for (scanf("%s%s", s + 1, ss + 1), n = strlen(s + 1), m = strlen(ss + 1),
                                     i = 0;
       i <= n; ++i)
    for (j = i + 1; j <= n; ++j) dp[i][j] = -inf;
  for (i = 1; i <= n; ++i)
    for (x = cal(i), j = 0; j <= n; ++j) {
      dp[i][j] = max(dp[i][j], dp[i - 1][j]);
      if (j >= x) dp[i][j] = max(dp[i][j], dp[i - x - m][j - x] + 1);
    }
  for (i = 0; i <= n; ++i) printf("%d ", dp[n][i]);
}
