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;
int dp[2000][2001];
string s, s2;
int main() {
  cin >> s;
  cin >> s2;
  if (s.size() < s2.size()) {
    for (int i = 0; i <= s.size(); ++i) printf("0 ");
    return 0;
  }
  for (int i = 0; i < s.size(); ++i) {
    dp[i][0] = (i > 0 ? dp[i - 1][0] : 0);
    for (int j = 1; j <= i; ++j) {
      if (i == 0) break;
      dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - 1]);
    }
    int fi = i;
    int se = s2.size() - 1;
    for (; fi >= 0 && se >= 0;) {
      if (s[fi] == s2[se]) {
        --fi;
        --se;
      } else
        --fi;
    }
    if (se == -1) {
      for (int j = 0; j <= fi + 1; ++j) {
        dp[i][i - fi - s2.size() + j] =
            max(dp[i][i - fi - s2.size() + j], 1 + (fi >= 0 ? dp[fi][j] : 0));
      }
    }
  }
  for (int i = 0; i <= s.size(); ++i) {
    printf("%d ", dp[s.size() - 1][i]);
  }
  return 0;
}
