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.*;
import java.io.*;

import static java.lang.Math.*;

public class C {
    public void run() throws IOException {
        in = new FastScanner(System.in);
        out = new PrintWriter(System.out);

        char[] s = in.next().toCharArray();
        char[] p = in.next().toCharArray();
        int n = s.length;
        int m = p.length;

        int[][] dp = new int[n + 1][n + 1];
        for (int[] z : dp) {
            Arrays.fill(z, Integer.MAX_VALUE);
        }
        dp[0][0] = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (dp[i][j] == Integer.MAX_VALUE) {
                    continue;
                }
                int k = j % m;
                if (s[i] == p[k]) {
                    dp[i + 1][j + 1] = Math.min(dp[i + 1][j + 1], dp[i][j]);
                }
                int delta = 1;
                if (j % m == 0) {
                    delta = 0;
                }
                dp[i + 1][j] = Math.min(dp[i + 1][j], dp[i][j] + delta);
            }
        }

        int[] minRem = new int[n / m + 2];
        Arrays.fill(minRem, Integer.MAX_VALUE);
        minRem[0] = 0;
        for (int i = 1; i <= n / m; i++) {
            for (int j = 0; j <= n; j++) {
                minRem[i] = Math.min(minRem[i], dp[j][i * m]);
            }
        }

        int v = 0;
        for (int x = 0; x <= n; x++) {
            while (minRem[v + 1] <= x) {
                v++;
            }
            out.print(Math.min(v, (n - x) / m));
            if (x < n) {
                out.print(" ");
            }
        }
        out.println();

        out.close();
    }

    FastScanner in;
    PrintWriter out;

    public static void main(String[] args) throws IOException {
        new C().run();
    }

    static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        FastScanner(InputStream in) throws IOException {
            br = new BufferedReader(new InputStreamReader(in));
        }

        String next() {
            try {
                while (st == null || !st.hasMoreTokens()) {
                    st = new StringTokenizer(br.readLine());
                }
            } catch (IOException e) {
                System.err.println(e);
                System.exit(1);
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }
    }
}