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", ""}.
#include <bits/stdc++.h>
using namespace std;
int N, M;
int dp[2005][2005], go[2005];
char ss[2005], p[2005];
int main() {
  int i, j;
  scanf("%s %s", &ss, &p);
  N = strlen(ss), M = strlen(p);
  for (i = 1; i <= N + 1; i++)
    for (j = 1; j <= M + 1; j++) dp[i][j] = 0x7ffffff;
  for (i = N; i >= 1; i--) {
    if (ss[i - 1] == p[M - 1]) dp[i][M] = i;
    for (j = M; j >= 1; j--)
      dp[i][j] = min(dp[i][j], ((ss[i - 1] == p[j - 1]) ? dp[i + 1][j + 1]
                                                        : dp[i + 1][j]));
    go[i] = dp[i][1];
  }
  memset(dp, 0, sizeof(dp));
  for (i = 0; i < N; i++)
    for (j = 0; j <= N; j++) {
      dp[i + 1][j] = max(dp[i + 1][j], dp[i][j]);
      if (j + go[i + 1] - i - M <= N)
        dp[go[i + 1]][j + go[i + 1] - i - M] =
            max(dp[go[i + 1]][j + go[i + 1] - i - M], dp[i][j] + 1);
    }
  for (i = 0; i <= N; i++) printf("%d ", min(dp[N][i], (N - i) / M));
  return 0;
}
