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.InputStreamReader;
import java.io.IOException;
import java.util.Arrays;
import java.io.BufferedReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
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();
	}
}

class TaskC {
    public void solve(int testNumber, InputReader in, PrintWriter out) {
        String s = in.next();
        String p = in.next();
        int n = s.length();
        int[][] last = new int[n][26];
        Arrays.fill(last[0], -1);
        last[0][s.charAt(0) - 'a'] = 0;
        for (int i = 1; i < n; ++i) {
            System.arraycopy(last[i - 1], 0, last[i], 0, 26);
            last[i][s.charAt(i) - 'a'] = i;
        }
        int[] closet = new int[n];
        outer: for (int end = 0; end < n; ++end) {
            int ptr = end + 1;
            for (int i = p.length() - 1; i >= 0; --i) {
                ptr = ptr == 0 ? -1 : last[ptr - 1][p.charAt(i) - 'a'];
                if (ptr == -1) {
                    closet[end] = -2;
                    continue outer;
                }
            }
            closet[end] = --ptr;
        }
        int[][] f = new int[n + 1][n + 1];
        Arrays.fill(f[0], -1000000000);
        f[0][0] = 0;
        for (int i = 1; i <= n; ++i)
            for (int j = 0; j <= n; ++j) {
                f[i][j] = f[i - 1][j];
                if (j > 0) f[i][j] = Math.max(f[i][j], f[i - 1][j - 1]);
                if (closet[i - 1] != -2) {
                    int needFix = (i - 1) - closet[i - 1] - p.length();
                    if (needFix <= j) f[i][j] = Math.max(f[i][j], f[closet[i - 1] + 1][j - needFix] + 1);
                }
            }
        for (int i = 0; i <= n; ++i) {
            if (i > 0) out.print(' ');
            out.print(f[n][i]);
        }
        out.println();
    }
}

class InputReader {
    private BufferedReader reader;
    private StringTokenizer tokenizer;

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

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

}

