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

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskC solver = new TaskC();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskC {
        public void solve(int testNumber, InputReader in, PrintWriter out) {
            char[] s = in.next().toCharArray(), p = in.next().toCharArray();
            int n = s.length, m = p.length;
            int[] end = new int[n];
            for (int i = 0; i < n; i++) {
                int j = i, k = 0;
                while (j < n && k < m) {
                    if (s[j] == p[k]) {
                        k++;
                    }
                    j++;
                }
                if (k == m) {
                    end[i] = j - 1;
                } else {
                    end[i] = -1;
                }
            }
            int[][] dp = new int[n][n + 1];
            if (end[0] != -1) {
                dp[end[0]][end[0] + 1 - m] = 1;
            }
            for (int i = 1; i < n; i++) {
                int e = end[i], l = end[i] - i + 1 - m;
                for (int j = 0; j <= i; j++) {
                    dp[i][j] = Math.max(dp[i][j], dp[i - 1][j]);
                    dp[i][j + 1] = Math.max(dp[i][j + 1], dp[i - 1][j]);
                    if (e != -1) {
                        dp[e][j + l] = Math.max(dp[e][j + l], dp[i - 1][j] + 1);
                    }
                }
            }
            for (int i = 0; i <= n; i++) {
                out.print(dp[n - 1][i] + " ");
            }
        }

    }

    static class InputReader {
        private BufferedReader br;
        private StringTokenizer st;

        public InputReader(InputStream stream) {
            br = new BufferedReader(new InputStreamReader(stream), 32768);
            st = null;
        }

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

    }
}

