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>
const long double eps = 1e-9;
const double pi = acos(-1.0);
const long long inf = 1e18;
using namespace std;
int n, m;
char s[2020], p[505];
int _next[2020];
int f[2020][2020], answer[2020];
void solve() {
  scanf("%s%s", s + 1, p + 1);
  n = strlen(s + 1);
  m = strlen(p + 1);
  for (int i = 1; i <= n; i++) {
    int pos = i;
    for (int j = 1; j <= m; j++) {
      while (true) {
        if (pos == n + 1) break;
        if (s[pos] == p[j]) break;
        pos += 1;
      }
      if (pos == n + 1) {
        pos = -1;
        break;
      }
      pos += 1;
    }
    if (pos == -1) {
      _next[i] = -1;
      continue;
    }
    _next[i] = pos - 1;
  }
  for (int i = 0; i < 2020; i++)
    for (int j = 0; j < 2020; j++) f[i][j] = -1;
  f[0][0] = 0;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      if (f[i][j] == -1) continue;
      f[i + 1][j] = max(f[i + 1][j], f[i][j]);
      f[i + 1][j + 1] = max(f[i + 1][j + 1], f[i][j]);
      if (_next[i + 1] != -1) {
        int rem = _next[i + 1] - i - m;
        if (j + rem <= n)
          f[_next[i + 1]][j + rem] = max(f[_next[i + 1]][j + rem], f[i][j] + 1);
      }
    }
  for (int i = 0; i <= n; i++) cout << f[n][i] << " ";
  cout << "\n";
}
int main(int argc, const char* argv[]) {
  time_t start = clock();
  int t = 1;
  for (int i = 0; i < t; i++) solve();
  cerr << fixed << setprecision(6)
       << "Time: " << 1.0 * (clock() - start) / CLOCKS_PER_SEC << "\n";
  return 0;
}
