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;
int main() {
  ios::sync_with_stdio(false);
  string s1, s2;
  cin >> s1 >> s2;
  int l1 = s1.length();
  int l2 = s2.length();
  int match[l1];
  for (int(i) = 0; (i) < (l1); (i)++) match[i] = -1;
  for (int(i) = 0; (i) < (l1); (i)++) {
    int cnt = 0;
    int id = 0;
    for (int(j) = (i); (j) < (l1); (j)++) {
      if (s1[j] == s2[id]) {
        id += 1;
        if (id == l2) {
          match[i] = j;
          break;
        }
      }
    }
  }
  int dp[l1 + 1][l1 + 1];
  memset(dp, (0), sizeof(dp));
  for (int(i) = 0; (i) < (l1); (i)++)
    for (int(j) = 0; (j) < (i + 1); (j)++) {
      dp[i + 1][j] = max(dp[i + 1][j], dp[i][j]);
      dp[i + 1][j + 1] = max(dp[i + 1][j + 1], dp[i][j]);
      if (match[i] != -1)
        dp[match[i] + 1][j + (1 + match[i] - i) - l2] =
            max(dp[match[i]][j + (1 + match[i] - i) - l2], 1 + dp[i][j]);
    }
  for (int(i) = 0; (i) < (l1 + 1); (i)++) cout << dp[l1][i] << " ";
  return 0;
}
