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 st1[2020], p[520];
int pre[2020], l1, l2;
int dp[2020][2020];
int main() {
  int i, j, k;
  while (scanf("%s", &st1) != EOF) {
    scanf("%s", &p);
    memset(pre, -1, sizeof(pre));
    l1 = strlen(st1);
    l2 = strlen(p);
    for (i = 0; i < l1; i++)
      if (st1[i] == p[l2 - 1]) {
        j = l2 - 1;
        for (k = i; k >= 0; k--)
          if (st1[k] == p[j]) {
            j--;
            if (j == -1) {
              pre[i] = k;
              break;
            }
          }
      }
    memset(dp, 0, sizeof(dp));
    for (i = 0; i <= l1; i++)
      for (j = i + 1; j <= l1; j++) {
        dp[j][i] = max(dp[j][i], max(dp[j - 1][i], dp[j - 1][i - 1]));
        if (pre[j - 1] != -1) {
          int t1 = j - pre[j - 1] - l2, t2 = i - t1;
          if (t1 <= i && pre[j - 1] - t2 >= 0)
            dp[j][i] = max(dp[j][i], dp[pre[j - 1]][t2] + 1);
        }
      }
    for (i = 0; i <= l1; i++) printf("%d ", dp[l1][i]);
  }
  return 0;
}
