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.*;
public class Test {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        char[] s = in.next().toCharArray();
        char[] p = in.next().toCharArray();
        int n = s.length;
        int m = p.length;
        int[][] dp = new int[n + 1][n + 1];
        for (int[] z : dp) {
            Arrays.fill(z, Integer.MAX_VALUE);
        }
        dp[0][0] = 0;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (dp[i][j] == Integer.MAX_VALUE) {
                    continue;
                }
                int k = j % m;
                if (s[i] == p[k]) {
                    dp[i + 1][j + 1] = Math.min(dp[i + 1][j + 1], dp[i][j]);
                }
                int delta = 1;
                if (j % m == 0) delta = 0;
                dp[i + 1][j] = Math.min(dp[i + 1][j], dp[i][j] + delta);
            }
        }
        int[] minRem = new int[n / m + 2];
        Arrays.fill(minRem, Integer.MAX_VALUE);
        minRem[0] = 0;
        for (int i = 1; i <= n / m; ++i) {
            for (int j = 0; j <= n; ++j) {
                minRem[i] = Math.min(minRem[i], dp[j][i * m]);
            }
        }
        int v = 0;
        for (int x = 0; x <= n; ++x) {
            while(minRem[v + 1] <= x) ++v;
            System.out.print(Math.min(v, (n - x) / m));
            if (x < n) System.out.print(" ");
        }
        System.out.println();
    }
}