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[1 << 11];
char S[1 << 11];
int A[1 << 11];
int R[1 << 11][1 << 11];
int main() {
  scanf("%s%s", S, P);
  int lenS = strlen(S);
  int lenP = strlen(P);
  int i, j;
  memset(A, -1, sizeof(A));
  for (i = (0); i < (lenS); ++i)
    if (S[i] == P[0]) {
      int cnt = 1;
      for (j = (i + 1); j < (lenS); ++j) {
        if (cnt == lenP) break;
        if (S[j] == P[cnt]) ++cnt;
      }
      if (cnt == lenP) A[i] = j;
    }
  for (i = (0); i < (lenS + 1); ++i)
    for (j = (0); j < (lenS + 1); ++j) R[i][j] = (1 << 30);
  R[0][0] = 0;
  for (i = (0); i < (lenS); ++i)
    for (j = (0); j < (lenS + 1); ++j)
      if (R[i][j] != (1 << 30)) {
        R[i + 1][j] = min(R[i + 1][j], R[i][j]);
        if (A[i] != -1) {
          int i2 = A[i];
          int extra = A[i] - i - lenP;
          R[i2][j + 1] = min(R[i2][j + 1], R[i][j] + extra);
        }
      }
  vector<int> best(lenS + 1, 0);
  for (i = (0); i < (lenS + 1); ++i)
    if (R[lenS][i] != (1 << 30)) best[R[lenS][i]] = max(best[R[lenS][i]], i);
  for (i = (0); i < (lenS); ++i) best[i + 1] = max(best[i + 1], best[i]);
  for (i = (0); i < (lenS + 1); ++i) {
    int t = min(best[i], (lenS - i) / lenP);
    printf("%d%c", t, i == lenS ? '\n' : ' ');
  }
  return 0;
};
