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 INF = 1000000000;
char s[2001], p[501];
int main() {
  scanf("%s", s);
  scanf("%s", p);
  int S = strlen(s), P = strlen(p);
  vector<vector<int> > dp(S + 1, vector<int>(S + 1, -INF));
  dp[S][0] = 0;
  for (int i = (S)-1; i >= 0; i--) {
    for (int j = (0); j <= (S); j++) {
      dp[i][j] = dp[i + 1][j];
    }
    for (int j = 0; j < (S); j++)
      dp[i][j + 1] = max(dp[i][j + 1], dp[i + 1][j]);
    int pp = 0, ss = i;
    while (pp < P) {
      while (ss < S && s[ss] != p[pp]) {
        ss++;
      }
      if (ss == S) {
        break;
      }
      ss++;
      pp++;
    }
    if (pp == P) {
      for (int j = (ss - i - P); j <= (S); j++)
        dp[i][j] = max(dp[i][j], 1 + dp[ss][j - (ss - i - P)]);
    }
  }
  for (int i = (0); i <= (S); i++) {
    if (i < S)
      printf("%d ", dp[0][i]);
    else
      printf("%d\n", dp[0][i]);
  }
  return 0;
}
