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;
static string s, p;
static int S, P;
static int skip_for[2001];
static int dp[2000][2001];
void look_for_pattern(int start) {
  int i = start, pct = 0;
  while (i < S) {
    if (s[i] == p[pct]) {
      pct++;
      i++;
      if (pct == P) {
        skip_for[start] = i - start - P;
        return;
      }
    } else
      i++;
  }
  skip_for[start] = -1;
}
int solve(int pos, int nskip) {
  if (S - pos - nskip < P || nskip < 0) return 0;
  if (dp[pos][nskip] >= 0) return dp[pos][nskip];
  if (skip_for[pos] > -1 && skip_for[pos] <= nskip)
    dp[pos][nskip] =
        max(max(solve(pos + 1, nskip), solve(pos + 1, nskip - 1)),
            solve(pos + P + skip_for[pos], nskip - skip_for[pos]) + 1);
  else
    dp[pos][nskip] = max(solve(pos + 1, nskip), solve(pos + 1, nskip - 1));
  return dp[pos][nskip];
}
int main() {
  cin >> s >> p;
  S = s.size();
  P = p.size();
  for (int i = 0; i < S; i++) look_for_pattern(i);
  for (int i = 0; i < S; i++)
    for (int j = 0; j < S + 1; j++) dp[i][j] = -1;
  for (int i = 0; i < S; i++) cout << solve(0, i) << ' ';
  cout << solve(0, S) << '\n';
  return 0;
}
