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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

/**
 * @author Don Li
 */
public class DreamoonStrings {

    static final int INF = (int) 1e5;

    void solve() {
        char[] s = in.nextToken().toCharArray();
        char[] p = in.nextToken().toCharArray();
        int n = s.length, m = p.length;

        int[] L = new int[n];
        for (int i = 0; i < n; i++) {
            int x = i, y = 0;
            while (x < n && y < m) {
                if (s[x] == p[y]) {
                    x++;
                    y++;
                } else {
                    x++;
                }
            }
            if (y == m) L[i] = x - i;
            else L[i] = INF;
        }

        int[][] dp = new int[n + 1][n + 1];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j <= i; j++) {
                dp[i + 1][j] = Math.max(dp[i + 1][j], dp[i][j]);
                dp[i + 1][j + 1] = Math.max(dp[i + 1][j + 1], dp[i][j]);
                if (L[i] < INF) dp[i + L[i]][j + L[i] - m] = Math.max(dp[i + L[i]][j + L[i] - m], dp[i][j] + 1);
            }
        }

        for (int j = 0; j <= n; j++) {
            if (j > 0) out.print(' ');
            out.print(dp[n][j]);
        }
        out.println();
    }

    public static void main(String[] args) {
        in = new FastScanner(new BufferedReader(new InputStreamReader(System.in)));
        out = new PrintWriter(System.out);
        new DreamoonStrings().solve();
        out.close();
    }

    static FastScanner in;
    static PrintWriter out;

    static class FastScanner {
        BufferedReader in;
        StringTokenizer st;

        public FastScanner(BufferedReader in) {
            this.in = in;
        }

        public String nextToken() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(in.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(nextToken());
        }

        public long nextLong() {
            return Long.parseLong(nextToken());
        }

        public double nextDouble() {
            return Double.parseDouble(nextToken());
        }
    }
}
