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;
char P[510];
char S[2010];
int N, M;
int len[2010];
int en[2010];
int mem[2010][2010];
void init() {
  int i, k;
  for (i = 0; i < N; ++i) {
    k = 0;
    en[i] = i;
    while (en[i] < N && k < M) {
      if (P[k] ^ S[en[i]]) {
        ++len[i];
        ++en[i];
        continue;
      }
      ++en[i];
      ++k;
    }
    en[i] = k < M ? -7 : en[i];
  }
}
int max_occ(int i, int k) {
  if (i < 0 || k < 0) return -(1e4);
  if (i >= N) return k ? -(1e4) : 0;
  if (mem[i][k] ^ -1) return mem[i][k];
  mem[i][k] = max(max_occ(en[i], k - len[i]) + 1, max_occ(i + 1, k));
  mem[i][k] = max(mem[i][k], max_occ(i + 1, k - 1));
  return mem[i][k];
}
int main() {
  int i;
  scanf("%s", S);
  scanf("%s", P);
  N = strlen(S);
  M = strlen(P);
  memset(mem, -1, sizeof(mem));
  init();
  for (i = 0; i <= N; ++i) {
    printf("%d ", max_occ(0, i));
  }
  return 0;
}
