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", ""}.
import java.util.*;
public class Test {
    static final int INF = (int) 1e9;
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        String s = in.next();
        String p = in.next();
        int plen = p.length();
        int[] len = new int[s.length()];
        Arrays.fill(len, INF);
        for (int i = 0; i + p.length() <= s.length(); ++i) {
            int l = 0;
            for (int j = 0; j < p.length(); ++j) {
                while (i + l < s.length() && s.charAt(i + l) != p.charAt(j)) ++l;
                if (i + l >= s.length()) {
                    l = INF;
                    break;
                }
                ++l;
            }
            len[i] = l;
        }
        int[][] best = new int[s.length() + 1][s.length() + 1];
        for (int[] x : best) Arrays.fill(x, INF);
        best[s.length()][0] = 0;
        for (int i = s.length() - 1; i >= 0; --i) {
            for (int j = 0; j <= s.length(); ++j) {
                best[i][j] = best[i + 1][j];
                if (j > 0 && len[i] < INF) best[i][j] = Math.min(best[i][j], best[i + len[i]][j - 1] + len[i] - plen);
            }
        }
        int[] res = new int[s.length() + 1];
        for (int take = 1; take <= s.length(); ++take) {
            int minDelete = best[0][take];
            int maxDelete = s.length() - take * plen;
            for (int x = minDelete; x <= maxDelete; ++x)
                res[x] = take;
        }
        for (int i = 0; i < res.length; ++i) {
            if (i > 0) System.out.print(' ');
            System.out.print(res[i]);
        }
        System.out.println();
    }
}