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;
#pragma GCC optimize("O2,no-stack-protector,unroll-loops,fast-math")
const int maxn = 2e3 + 10, maxm = 1e3 + 10, lg = 12, mod = 1e9 + 7, inf = 1e18;
string s, t;
int n, m, dp[maxn][maxn], id[maxn];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  memset(id, -1, sizeof(id));
  cin >> s >> t;
  n = (int)(s.size()), m = (int)(t.size());
  for (int i = 0; i < n; i++) {
    int nn = m - 1, j;
    if (s[i] != t[nn]) continue;
    for (j = i; j >= 0 && nn >= 0; j--)
      if (s[j] == t[nn]) nn--;
    if (nn == -1) id[i] = j + 1;
  }
  for (int i = 0; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      dp[i][j] = dp[i][j - 1];
      if (i == 0) {
        if (id[j - 1] != -1 && j - id[j - 1] == m)
          dp[0][j] = dp[0][id[j - 1]] + 1;
      } else if (id[j - 1] != -1) {
        int xx = i - (j - id[j - 1]) + m;
        if (xx >= 0) dp[i][j] = max(dp[i][j], dp[xx][id[j - 1]] + 1);
      }
    }
  for (int i = 0; i <= n; i++) cout << min((n - i) / m, dp[i][n]) << ' ';
  return 0;
}
