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 N = 2048;
char s[N];
char p[N];
int M[N][N];
int Z[N];
const int MX = 20000000;
int main() {
  scanf("%s", s);
  scanf("%s", p);
  int L = strlen(s);
  int PL = strlen(p);
  for (int i = 0; i <= L; ++i) Z[i] = MX;
  for (int i = 0; i < L - PL + 1; ++i) {
    int ip = 0;
    for (int j = i; j < L; ++j) {
      if (s[j] == p[ip]) {
        ++ip;
      }
      if (PL == ip) {
        Z[i] = j;
        break;
      }
    }
  }
  for (int i = L - 1; i >= 0; --i)
    if (Z[i] > Z[i + 1]) Z[i] = Z[i + 1];
  int C = L / PL;
  for (int i = 1; i <= C; ++i)
    for (int j = 0; j <= L; ++j) M[j][i] = MX;
  for (int i = L - 1; i >= 0; --i) {
    for (int j = 1; j <= C; ++j) {
      if (Z[i] < L) {
        M[i][j] = M[Z[i] + 1][j - 1] + Z[i] - i - PL + 1;
        M[i][j] = min(M[i][j], M[i + 1][j]);
      }
    }
  }
  for (int i = 0; i <= L; ++i) {
    int k = 0;
    for (int j = 0; j <= C; ++j) {
      if (M[0][j] <= i) k = j;
    }
    int rs = 0;
    if (L - k * PL < i) {
      rs = (i - L + k * PL) / PL;
      if ((i - L + k * PL) % PL) ++rs;
    }
    if (i < L)
      printf("%d ", k - rs);
    else
      printf("%d\n", k - rs);
  }
  return 0;
}
