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>
const int N = 2222, M = 555;
char S[N], P[M];
int n, m;
int dump[N][M], f[N][N];
bool solved[N][N];
__inline void cmin(int &f0, int f) {
  if (!~f0 || f0 > f) f0 = f;
}
int solve(int i, int j) {
  if (j < 0) return -1;
  if (solved[i][j]) return f[i][j];
  solved[i][j] = true;
  if (i == n) return f[i][j] = -(int)(j > 0);
  f[i][j] = std::max(solve(i + 1, j), solve(i + 1, j - 1));
  if (~dump[i][0]) {
    int tmp = solve(i + m + dump[i][0], j - dump[i][0]);
    if (~tmp) f[i][j] = std::max(f[i][j], tmp + 1);
  }
  return f[i][j];
}
int main() {
  std::cin >> S >> P;
  n = strlen(S);
  m = strlen(P);
  memset(dump, -1, sizeof dump);
  dump[n][m] = 0;
  for (int i = n - 1; i >= 0; --i) {
    dump[i][m] = 0;
    for (int j = 0; j < m; ++j) {
      if (S[i] == P[j] && ~dump[i + 1][j + 1])
        cmin(dump[i][j], dump[i + 1][j + 1]);
      if (~dump[i + 1][j]) cmin(dump[i][j], dump[i + 1][j] + 1);
    }
  }
  memset(solved, 0, sizeof solved);
  for (int i = 0; i <= n; ++i) std::cout << solve(0, i) << " ";
  std::cout << std::endl;
  return 0;
}
