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.util.Arrays;

import static java.lang.Math.max;

/**
 * ToDo: describe
 *
 * @author g.troitskiy
 */
public class E {
    public static void main(String[] args) throws IOException {
        final BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        final String s = reader.readLine();
        final String p = reader.readLine();
        final int n = s.length();
        final int pn = p.length();
        final int[][] mn = new int[n + 1][n + 1];
        final int[] last = new int[n + 1];
        Arrays.fill(last, -1);
        for (int i = 0; i <= n - pn; i++) {
            int k = 0;
            int j = i;
            while (j < n && k < pn) {
                if (s.charAt(j) == p.charAt(k)) {
                    k++;
                }
                j++;
            }
            if (k == pn) {
                last[i] = j - i;
            }
        }

        for (int k = 0; k <= n; k++) {
            for (int i = k; i <= n; i++) {
                if (i < n) {
                    mn[i + 1][k] = max(mn[i + 1][k], mn[i][k]);
                    if (k < n) mn[i + 1][k + 1] = max(mn[i + 1][k + 1], mn[i][k]);
                }
                int b = last[i];
                if (b == -1) continue;
                if (i + b <= n && k + (b - pn) <= n) {
                    mn[i + b][k + (b - pn)] = max(mn[i + b][k + (b - pn)], mn[i][k] + 1);
                }
            }
        }
        for (int i = 0; i <= n; i++) {
            if (i > 0) System.out.print(" ");
            System.out.print(mn[n][i]);
        }
    }
}