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 ne[2010];
int cost[2010];
int dp[2010][2010];
string s, p;
int solve(int curr, int qnt) {
  if (curr == s.size() && qnt == 0)
    return 0;
  else if (curr == s.size())
    return -0x3f3f3f3f;
  int& ret = dp[curr][qnt];
  if (ret != -1) return ret;
  ret = -0x3f3f3f3f;
  if (ne[curr] != -1) {
    if (qnt - cost[curr] >= 0)
      ret = max(ret, solve(ne[curr], qnt - cost[curr]) + 1);
  }
  ret = max(ret, solve(curr + 1, qnt));
  ret = max(ret, solve(curr + 1, qnt - 1));
  return ret;
}
int main(void) {
  cin >> s >> p;
  memset(ne, -1, sizeof ne);
  for (int i = 0; i < s.size(); ++i) {
    if (s[i] != p[0]) continue;
    int k = 1;
    int j = i + 1;
    int custo = 0;
    for (j = i + 1; j < s.size(); ++j) {
      if (k == p.size()) break;
      if (s[j] == p[k])
        k++;
      else
        custo++;
    }
    if (k == p.size()) {
      cost[i] = custo;
      ne[i] = j;
    }
  }
  memset(dp, -1, sizeof dp);
  for (int i = 0; i <= s.size(); ++i) {
    cout << solve(0, i);
    if (i == s.size())
      cout << "\n";
    else
      cout << " ";
  }
  return 0;
}
