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>
const int MX = 2222;
using namespace std;
char s1[MX], s2[MX];
int n, m, dp[MX][MX];
int nd(int i) {
  if (m > i) return MX;
  int top = m, ans = 0;
  if (i < m) return MX;
  while (top && i) {
    if (s1[i] == s2[top])
      --top;
    else
      ++ans;
    --i;
  }
  if (top)
    return MX;
  else
    return ans;
}
int main() {
  while (~scanf("%s%s", s1 + 1, s2 + 1)) {
    n = strlen(s1 + 1), m = strlen(s2 + 1);
    memset(dp, 0, sizeof(dp));
    for (int i = 0; i <= n; i++)
      for (int j = i + 1; j <= n; j++) dp[i][j] = -MX;
    for (int i = 1; i <= n; i++) {
      int k = nd(i);
      for (int j = 0; j <= i; j++) {
        dp[i][j] = max(dp[i - 1][j], dp[i][j]);
        if (i - m - k >= 0 && j >= k)
          dp[i][j] = max(dp[i][j], dp[i - m - k][j - k] + 1);
      }
    }
    for (int i = 0; i <= n; i++) printf("%d ", dp[n][i]);
    puts("");
  }
  return 0;
}
