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;
string s;
string p;
int N, K;
int dp[2010][2010];
int main(void) {
  int i, j;
  cin >> s >> p;
  N = s.length();
  K = p.length();
  for ((i) = 0; (i) < (int)(N + 1); (i)++)
    for ((j) = 0; (j) < (int)(N + 1); (j)++) dp[i][j] = (1 << 29);
  dp[0][0] = 0;
  for ((i) = 0; (i) < (int)(N); (i)++)
    for ((j) = 0; (j) < (int)(i + 1); (j)++)
      if (dp[i][j] != (1 << 29)) {
        if (p[j % K] == s[i])
          dp[i + 1][j + 1] = min(dp[i + 1][j + 1], dp[i][j]);
        int tmp = dp[i][j];
        if (j % K != 0) tmp++;
        dp[i + 1][j] = min(dp[i + 1][j], tmp);
      }
  for ((i) = 0; (i) < (int)(N + 1); (i)++) {
    for (j = N / K;; j--)
      if (dp[N][j * K] <= i) break;
    int ans = j;
    ans = min(ans, (N - i) / K);
    printf("%d", ans);
    if (i == N)
      printf("\n");
    else
      printf(" ");
  }
  return 0;
}
