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 INF = (1LL << 31) - 1;
const long long int LINF = (1LL << 62) - 1;
const int NMAX = 2000 + 5;
const int MMAX = 500 + 5;
const int KMAX = 100000 + 5;
const int PMAX = 100000 + 5;
const int LMAX = 100000 + 5;
const int VMAX = 100000 + 5;
int N, M;
char S[NMAX];
char P[MMAX];
int A[NMAX];
int DP[NMAX][NMAX];
int main() {
  int i, j, k, len;
  scanf("%s", S);
  scanf("%s", P);
  N = strlen(S);
  M = strlen(P);
  for (i = 0; i < N; i++) {
    for (k = 0, len = 0; k < M && i + len < N; len++)
      if (S[i + len] == P[k]) k++;
    if (k < M)
      A[i] = -1;
    else
      A[i] = len;
  }
  for (i = 0; i < N; i++) {
    for (j = 0; j <= i; j++) {
      DP[i + 1][j] = max(DP[i + 1][j], DP[i][j]);
      DP[i + 1][j + 1] = max(DP[i + 1][j + 1], DP[i][j]);
      if (A[i] >= 0)
        DP[i + A[i]][j + A[i] - M] =
            max(DP[i + A[i]][j + A[i] - M], DP[i][j] + 1);
    }
  }
  for (i = 0; i <= N; i++) printf("%d ", DP[N][i]);
  return 0;
}
