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 A[2010];
vector<int> B[2010];
int dp[2010][2010];
int main() {
  string a;
  string b;
  cin >> a >> b;
  for (int i = 0; i < a.length(); i++) {
    int curMatch = 0;
    for (int j = i; j < a.length(); j++) {
      if (a[j] == b[curMatch]) curMatch++;
      if (curMatch == b.length()) {
        A[1 + i] = 1 + j;
        B[1 + j].push_back(1 + i);
        break;
      }
    }
    if (curMatch != b.length()) A[1 + i] = 99999;
  }
  for (int deleted = 0; deleted < a.length(); deleted++) {
    if (deleted > 0) dp[deleted][0] = -99999;
    for (int i = 1; i <= a.length(); i++) {
      if (deleted > 0) dp[deleted][i] = dp[deleted - 1][i - 1];
      dp[deleted][i] = max(dp[deleted][i], dp[deleted][i - 1]);
      for (int j = 0; j < B[i].size(); j++) {
        int siz = i - B[i][j] + 1 - b.size();
        if (deleted - siz >= 0)
          dp[deleted][i] =
              max(dp[deleted][i], dp[deleted - siz][B[i][j] - 1] + 1);
      }
      if (deleted > i) dp[deleted][i] = -99999;
    }
  }
  int sol = 0;
  vector<int> ans;
  for (int i = a.length(); i >= 0; i--) {
    sol = max(0, dp[i][a.length()]);
    ans.push_back(sol);
  }
  reverse(ans.begin(), ans.end());
  for (int i = 0; i < ans.size(); i++) printf("%d ", ans[i]);
  printf("\n");
  return 0;
}
