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

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

class Task {
    public void solve(int testNumber, Reader in, PrintWriter out) {
        String s, p;
        s = in.rl(); p = in.rl();
        int n, m;
        n = s.length(); m = p.length();
        int[] t = new int[n];
        boolean[] found = new boolean[m];
        for(int i = 0; i < n; i++) {
            Arrays.fill(found, false);
            int j = i;
            int curLetter = 0;
            for(; j < n && curLetter < m; j++) {
                if(s.charAt(j) == p.charAt(curLetter)) {
                    found[curLetter++] = true;
                }
            }
            if(curLetter == m) {
                t[i] = j - i;
            } else {
                t[i] = -1;
            }
        }
        int[][] dp = new int[n + 1][n + 1];
        for(int i = 0; i <= n; i++) {
            for(int j = 0; j <= n; j++) {
                dp[i][j] = -1;
            }
        }
        dp[0][0] = 0;
        for(int i = 0; i < n; i++) {
            for(int j = 0; j <= n; j++) {
                if(dp[i][j] == -1) {
                    continue;
                }
                dp[i + 1][j] = Math.max(dp[i + 1][j], dp[i][j]);
                if(j + 1 <= n) {
                    dp[i + 1][j + 1] = Math.max(dp[i + 1][j + 1], dp[i][j]);
                }
                if(t[i] != -1) {
                    if(i + t[i] <= n && j + t[i] - m <= n) {
                        dp[i + t[i]][j + t[i] - m] = Math.max(dp[i + t[i]][j + t[i] - m], dp[i][j] + 1);
                    }
                }
            }
        }
        for(int i = 0; i <= n; i++) {
            out.print(dp[n][i] + " ");
        }

    }
}

class Reader {
    private BufferedReader in;

    public Reader(InputStream in) {
        this.in = new BufferedReader(new InputStreamReader(in));
    }

    public String rl() {
        try {
            return in.readLine();
        } catch(IOException e) {
            throw new RuntimeException(e);
        }
    }

}

