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 int MAXLS = 2005;
const int MAXLP = 505;
int ls, lp;
char s[MAXLS];
char p[MAXLP];
int matchend[MAXLS];
int f[MAXLS][MAXLS];
int g[MAXLS][MAXLS];
int cg[MAXLS];
void init() {
  scanf("%s%s", s, p);
  ls = strlen(s);
  lp = strlen(p);
  for (int i = 0; i < ls; ++i) {
    int j = i;
    for (int k = 0; k < lp; ++k) {
      while (j < ls && s[j] != p[k]) ++j;
      ++j;
    }
    if (j > ls)
      matchend[i] = -1;
    else
      matchend[i] = j;
  }
}
void solve() {
  for (int i = 0; i <= ls; ++i) {
    for (int j = 0; j <= i; ++j) {
      if (g[i][j] > cg[j]) cg[j] = g[i][j];
      if (cg[j] > f[i][j]) f[i][j] = cg[j];
    }
    if (i == ls) break;
    for (int j = 0; j <= i; ++j) {
      if (f[i][j] > f[i + 1][j + 1]) f[i + 1][j + 1] = f[i][j];
      if (matchend[i] != -1 && f[i][j] + 1 > g[matchend[i]][j + lp])
        g[matchend[i]][j + lp] = f[i][j] + 1;
    }
  }
  for (int i = ls; i >= 0; --i) printf("%d ", f[ls][i]);
  printf("\n");
}
int main() {
  init();
  solve();
  return 0;
}
