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 s1[2010], s2[555];
int len1, len2;
int dp[2010][2010];
int pre[2010];
int main() {
  scanf("%s%s", s1 + 1, s2 + 1);
  len1 = strlen(s1 + 1);
  len2 = strlen(s2 + 1);
  memset(pre, 0, sizeof(pre));
  for (int i = len2; i <= len1; i++) {
    int now = len2;
    for (int j = i; j >= 1; j--)
      if (s1[j] == s2[now]) {
        now--;
        if (now == 0) {
          pre[i] = j;
          break;
        }
      }
  }
  for (int i = 0; i <= len1; i++)
    for (int j = 0; j <= len1; j++)
      if (j <= i)
        dp[i][j] = 0;
      else
        dp[i][j] = -99999999;
  for (int i = 1; i <= len1; i++) {
    if (pre[i] == 0) continue;
    for (int j = 0; j <= i; j++) {
      int res = i - pre[i] + 1 - len2;
      dp[i][j] = max(dp[i][j], dp[i - 1][j]);
      if (j > 0) dp[i][j] = max(dp[i][j], dp[i - 1][j - 1]);
      if (j >= res) dp[i][j] = max(dp[i][j], dp[pre[i] - 1][j - res] + 1);
    }
  }
  for (int i = 0; i <= len1; i++) {
    if (i == len1)
      printf("%d\n", dp[len1][i]);
    else
      printf("%d ", dp[len1][i]);
  }
  return 0;
}
