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;
const int maxn = 2005;
char s[maxn], p[maxn];
int _dp[maxn * maxn * 2];
int ans[maxn];
int ls, lp;
int& dp(int poss, int posp, int alrp) {
  return _dp[poss * 2 * maxn + alrp * lp + posp];
}
void updDp(int a, int b, int c, int d) {
  int& x = dp(a, b, c);
  if (x == -1 || x > d) {
    x = d;
  }
}
int main() {
  scanf("%s", s);
  scanf("%s", p);
  ls = strlen(s);
  lp = strlen(p);
  memset(_dp, -1, sizeof _dp);
  dp(0, 0, 0) = 0;
  int maxIn = ls / lp + 1;
  for (int i = 0; i < ls; i++)
    for (int j = 0; j < lp; j++)
      for (int alr = 0; alr < maxIn; alr++) {
        int cur = dp(i, j, alr);
        if (cur == -1) continue;
        if (s[i] == p[j]) {
          int ni = i + 1;
          int nj = j + 1;
          int nalr = alr;
          if (nj == lp) {
            nj = 0;
            nalr++;
          }
          updDp(ni, nj, nalr, cur);
        }
        if (j == 0) {
          int ni = i + 1;
          int nj = 0;
          int nalr = alr;
          updDp(ni, nj, nalr, cur);
        }
        int ni = i + 1;
        int nj = j;
        int nalr = alr;
        updDp(ni, nj, nalr, cur + 1);
      }
  for (int cnt = 0; cnt < maxIn; cnt++) {
    int minDel = dp(ls, 0, cnt);
    if (minDel == -1) continue;
    int maxDel = ls - cnt * lp;
    for (int i = minDel; i <= maxDel; i++) ans[i] = max(ans[i], cnt);
  }
  for (int i = 0; i <= ls; i++) {
    printf("%d ", ans[i]);
  }
  return 0;
}
