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;
string s, p;
int n, m, inf = 0x3f3f3f3f;
int h[2006][506], g[2006], f[2006][2006];
inline void umin(int& a, int b) {
  if (a > b) a = b;
}
inline void umax(int& a, int b) {
  if (a < b) a = b;
}
void goh() {
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) h[i][j] = inf;
  for (int i = n - 1; i >= 0; i--)
    for (int j = m - 1; j >= 0; j--) {
      if (s[i] != p[j] and i != n - 1) umin(h[i][j], 1 + h[i + 1][j]);
      if (s[i] == p[j] and !(i == n - 1 and j < m - 1))
        umin(h[i][j], h[i + 1][j + 1]);
    }
}
void gog() {
  for (int i = 0; i < n; i++) g[i] = h[i][0];
}
void gof() {
  for (int i = 0; i < 2006; i++)
    for (int j = 0; j < 2006; j++) f[i][j] = -inf;
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= n - i; j++) f[i][j] = 0;
  for (int i = n - 1; i >= 0; i--)
    for (int j = n - i; j >= 0; j--) {
      umax(f[i][j], f[i + 1][j]);
      if (j > 0) umax(f[i][j], f[i + 1][j - 1]);
      if (s[i] == p[0] and g[i] < inf and j >= g[i])
        umax(f[i][j], 1 + f[i + g[i] + m][j - g[i]]);
    }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> s >> p;
  n = s.size(), m = p.size();
  goh(), gog(), gof();
  for (int i = 0; i <= n; i++) cout << f[0][i] << " ";
  cout << "\n";
  return 0;
}
