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[10001], s2[10001];
int x[2002], l1, l2, dp[2002][2002];
int main() {
  scanf("%s%s", s1, s2);
  l1 = strlen(s1);
  l2 = strlen(s2);
  for (int i = 0; i < l1; i++) {
    int r = 0, j;
    for (j = i; j < l1 && r < l2; j++)
      if (s1[j] == s2[r]) r++;
    if (r == l2)
      x[i] = j;
    else
      x[i] = -1;
  }
  for (int i = 0; i <= l1; i++)
    for (int j = l1 - 1; j >= 0; j--) {
      dp[j][i] = dp[j + 1][i];
      if (i != 0) dp[j][i] = max(dp[j + 1][i - 1], dp[j][i]);
      if (x[j] != -1 && i - (x[j] - j - l2) >= 0 &&
          (l1 - x[j] >= i - (x[j] - j - l2)))
        dp[j][i] = max(dp[j][i], dp[x[j]][i - (x[j] - j - l2)] + 1);
    }
  for (int i = 0; i <= l1; i++) {
    int ans = 0;
    for (int j = 0; j < l1; j++) ans = max(ans, dp[j][i]);
    printf("%d ", ans);
  }
  return 0;
}
