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 dp[2000][2001];
int last[2000];
string s, p;
int calc(int i) {
  int j = p.size() - 1;
  for (; i >= 0; i--) {
    if (s[i] == p[j]) j--;
    if (j < 0) return i;
  }
  return -1;
}
int main() {
  cin >> s >> p;
  int N = s.size();
  int M = p.size();
  for (int i = 0; i < s.size(); i++) last[i] = calc(i);
  for (int i = 0; i < s.size(); i++) {
    for (int x = 0; x <= s.size(); x++) {
      dp[i][x] = -1;
      if (x > (i + 1)) continue;
      if (((i + 1 - last[i] - M) <= x) && ((i + 1 - M) >= x)) dp[i][x] = 1;
      if ((last[i] > 0) && (((i + 1 - last[i]) - M) <= x) &&
          (dp[last[i] - 1][x - ((i + 1 - last[i]) - M)] > -1))
        dp[i][x] = 1 + dp[last[i] - 1][x - ((i + 1 - last[i]) - M)];
      else
        dp[i][x] = max(dp[i][x], 0);
      if (i > 0) dp[i][x] = max(dp[i][x], dp[i - 1][x]);
    }
  }
  for (int x = 0; x <= s.size(); x++) cout << dp[s.size() - 1][x] << " ";
  return 0;
}
