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>
int f[2020][2020];
char s[2020];
char p[2020];
int g[2020][2020];
int pos[2020];
int n, m;
void maximize(int &a, int b) {
  if (a < b) a = b;
}
void read_precal() {
  scanf("%s", &s);
  scanf("%s", &p);
  n = strlen(s);
  m = strlen(p);
  for (int i = 1; i <= n; i++) {
    int curr = 1;
    for (int j = i; j <= n; j++) {
      g[i][j] = -1;
      if (curr <= m && s[j - 1] == p[curr - 1]) curr++;
      if (curr == m + 1) g[i][j] = 1;
    }
  }
}
void optimize() {
  for (int i = 1; i <= n; i++) {
    bool ok = false;
    for (int j = i; j >= 1; j--)
      if (g[j][i] != -1) {
        pos[i] = j;
        ok = true;
        break;
      }
    if (!ok) pos[i] = -1;
  }
}
void solve() {
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= n; j++) f[i][j] = -10000000;
  f[0][0] = 0;
  for (int j = 0; j <= n; j++)
    for (int i = 1; i <= n; i++) {
      maximize(f[i][j], f[i - 1][j]);
      if (j >= 1) maximize(f[i][j], f[i - 1][j - 1]);
      if (pos[i] != -1 && (i - pos[i] + 1 - m) <= j)
        maximize(f[i][j], f[pos[i] - 1][j - (i - pos[i] + 1 - m)] + 1);
    }
  for (int i = 0; i <= n; i++) printf("%d ", f[n][i]);
}
int main() {
  read_precal();
  optimize();
  solve();
}
