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;
long mb[2002][502];
long dp[2002][2002];
int main() {
  string s, p;
  cin >> s >> p;
  int n = (int)s.length();
  for (int i = 0; i < n; i++) mb[i][0] = 0;
  for (int i = 1; i <= p.length(); i++) {
    for (int j = 0; j < n; j++) {
      if (j < i - 1) {
        mb[j][i] = -1;
        continue;
      }
      if (s[j] == p[i - 1]) {
        if (mb[j - 1][i - 1] == -1)
          mb[j][i] = -1;
        else if (j > 0)
          mb[j][i] = mb[j - 1][i - 1] + 1;
        else
          mb[j][i] = 1;
      } else {
        if (mb[j - 1][i] == -1)
          mb[j][i] = -1;
        else if (j > 0)
          mb[j][i] = mb[j - 1][i] + 1;
        else
          mb[j][i] = -1;
      }
    }
  }
  int m = (int)p.size();
  for (int i = 0; i < n; i++) {
    if (i < m - 1) {
      dp[i][0] = 0;
      continue;
    }
    int b = 0;
    for (int j = i; j > i - m; j--) {
      if (s[j] != p[j - i + m - 1]) {
        b = 1;
        break;
      }
    }
    if (!b) {
      dp[i][0] = 1;
      if (i - m >= 0) dp[i][0] += dp[i - m][0];
    } else if (i) {
      dp[i][0] = dp[i - 1][0];
    }
  }
  for (int i = 1; i < n; i++) {
    for (int j = 1; j <= i + 1; j++) {
      if (i + 1 - j < m)
        dp[i][j] = 0;
      else {
        dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - 1]);
        if (mb[i][m] != -1 && i >= mb[i][m] - 1 && j >= mb[i][m] - m) {
          if (i >= mb[i][m])
            dp[i][j] = max(dp[i][j], dp[i - mb[i][m]][j - mb[i][m] + m] + 1);
          else
            dp[i][j] = max(dp[i][j], (long)1);
        }
      }
    }
  }
  for (int i = 0; i <= n; i++) {
    cout << dp[n - 1][i] << ' ';
  }
  cout << endl;
}
