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 s[2005], p[505];
int dp[2005][2005], fir[2005], lens, lenp;
int getf(int ed) {
  int i = ed, j = lenp - 1;
  while (j >= 0) {
    while (i >= 0 && s[i] != p[j]) i--;
    if (i < 0) return -1;
    i--, j--;
  }
  return i + 1;
}
int main() {
  scanf("%s%s", s, p);
  lens = strlen(s);
  lenp = strlen(p);
  for (int i = 0; s[i]; i++) {
    fir[i] = getf(i);
  }
  if (fir[0] != -1) dp[0][0] = 1;
  for (int i = 1; s[i]; i++) {
    for (int j = 0; j <= i + 1; j++) dp[i][j] = dp[i - 1][j];
    if (fir[i] != -1) {
      for (int j = i + 1 - fir[i] - lenp;
           j <= i + 1 && j - i - 1 + fir[i] + lenp <= fir[i]; j++) {
        if (fir[i] == 0) {
          dp[i][j] = max(dp[i][j], 1);
        } else {
          dp[i][j] =
              max(dp[i][j], dp[fir[i] - 1][j - i - 1 + fir[i] + lenp] + 1);
        }
      }
    }
  }
  for (int i = 0; i <= lens; i++) {
    printf("%d ", dp[lens - 1][i]);
  }
  puts("");
  return 0;
}
