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 C[2005], S[505];
int n, m, x, y, z, k, w, A[2005][505], B[505];
int s[2005], e[2005], T[2005], DP[2005][2005];
int main() {
  scanf("%s", &C);
  scanf("%s", &S);
  n = strlen(C);
  m = strlen(S);
  memset(B, -1, sizeof(B));
  memset(A, -1, sizeof(A));
  for (int i = (0); i < (n); i++) {
    for (int j = (m - 1); j >= (0); j--)
      if (C[i] == S[j]) {
        if (j == 0) {
          A[i][j] = i;
          B[j] = i;
        } else {
          A[i][j] = B[j - 1];
          B[j] = B[j - 1];
        }
      }
  }
  for (int i = (0); i < (n); i++) {
    if (A[i][m - 1] != -1 && (A[i][m - 1] != s[w - 1] || w == 0)) {
      s[w] = A[i][m - 1];
      e[w++] = i;
    }
  }
  memset(T, -1, sizeof(T));
  for (int i = (0); i < (w); i++) T[e[i] + 1] = s[i] + 1;
  memset(DP, -1, sizeof(DP));
  for (int i = (0); i <= (n); i++) DP[0][i] = 0;
  for (int i = (1); i <= (n); i++) {
    for (int j = (1); j <= (n); j++) {
      if (T[j] != -1 && DP[i - 1][T[j] - 1] != -1)
        DP[i][j] = DP[i - 1][T[j] - 1] + j - T[j] + 1 - m;
      if (DP[i][j] == -1 && j > 1 && DP[i][j - 1] != -1)
        DP[i][j] = DP[i][j - 1];
      else if (j > 1 && DP[i][j - 1] != -1)
        DP[i][j] = min(DP[i][j], DP[i][j - 1]);
    }
  }
  for (int i = (1); i <= (n); i++)
    if (DP[i][n] != -1) y = i;
  x = 0;
  z = 0;
  for (int i = (0); i <= (n); i++) {
    if (i > n - y * m) {
      if (k % m == 0) z++;
      k++;
    }
    while (DP[x + 1][n] <= i && x + 1 <= n && DP[x + 1][n] != -1) x++;
    if (i != n)
      printf("%d ", x - z);
    else
      printf("%d\n", x - z);
  }
  return 0;
}
