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;
const int mod = 1e9 + 7;
const int MAXN = 2010;
int l[MAXN], cnt[MAXN];
int dp[MAXN][MAXN];
int main() {
  string s, p;
  cin >> s >> p;
  for (int i = 0; i < s.size(); i++) {
    int ind = p.size() - 1;
    int ind_s = i;
    cnt[i + 1] = 0;
    while (ind >= 0 && ind_s >= 0) {
      if (s[ind_s] == p[ind])
        ind--;
      else
        cnt[i + 1]++;
      ind_s--;
    }
    if (ind < 0)
      l[i + 1] = ind_s + 1;
    else
      l[i + 1] = -1;
  }
  dp[0][0] = 0;
  for (int i = 1; i <= s.size(); i++) dp[0][i] = -1000000;
  for (int i = 1; i <= s.size(); i++) {
    int ii = i;
    for (int j = 0; j <= s.size(); j++) {
      dp[ii][j] = dp[ii - 1][j];
      if (j) dp[ii][j] = max(dp[ii][j], dp[ii - 1][j - 1]);
      int L = l[i];
      int Cnt = cnt[i];
      if (L >= 0 && Cnt <= j) {
        dp[ii][j] = max(dp[ii][j], dp[L][j - Cnt] + 1);
      }
    }
  }
  for (int i = 0; i <= s.size(); i++) cout << dp[s.size()][i] << " ";
  return 0;
}
