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;
const int maxn = 3005;
char s[maxn], p[maxn];
int ls, lp, r[maxn];
int dp[maxn][maxn];
int calc(int x) {
  int y = 1;
  for (int i = x; i <= ls; i++) {
    if (s[i] == p[y]) y++;
    if (y == lp + 1) return i;
  }
  return -1;
}
int main() {
  scanf("%s%s", s + 1, p + 1);
  ls = strlen(s + 1);
  lp = strlen(p + 1);
  for (int i = 1; i <= ls; i++) {
    r[i] = calc(i);
  }
  for (int i = 0; i < ls; i++) {
    for (int j = 0; j <= i; j++) {
      if (s[i + 1] == p[1] && ~r[i + 1]) {
        dp[r[i + 1]][j + (r[i + 1] - (i + 1) + 1 - lp)] =
            max(dp[r[i + 1]][j + (r[i + 1] - (i + 1) + 1 - lp)], dp[i][j] + 1);
      }
      if (j < ls) dp[i + 1][j + 1] = max(dp[i + 1][j + 1], dp[i][j]);
      dp[i + 1][j] = max(dp[i + 1][j], dp[i][j]);
    }
  }
  for (int i = 0; i <= ls; i++) printf("%d%c", dp[ls][i], i == ls ? '\n' : ' ');
  return 0;
}
