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;
char p[510];
char s[2100];
int dp[2100][2100];
int nex[2100], len[2100];
int main() {
  scanf("%s", &s[1]);
  scanf("%s", &p[1]);
  int n = strlen(&s[1]), m = strlen(&p[1]);
  for (int i = 1; i <= n; ++i) {
    int j = i - 1;
    for (int k = 1; k <= m && j <= n; ++k) {
      ++j;
      while (j <= n && s[j] != p[k]) ++j;
    }
    nex[i] = j;
    len[i] = j - i + 1;
  }
  puts("");
  puts("");
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j <= i; ++j) {
      dp[i + 1][j + 1] = max(dp[i][j], dp[i + 1][j + 1]);
      dp[i + 1][j] = max(dp[i][j], dp[i + 1][j]);
      if (nex[i + 1] <= n)
        dp[nex[i + 1]][len[i + 1] - m + j] =
            max(dp[nex[i + 1]][len[i + 1] - m + j], dp[i][j] + 1);
    }
  }
  for (int i = 0; i < n; ++i) printf("%d ", dp[n][i]);
  printf("%d\n", dp[n][n]);
  return 0;
}
