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;
int nextm[2005];
string s, p;
int calcNext(int si, int pi) {
  if (pi == p.size()) return si;
  if (si == s.size()) return -1;
  if (s[si] == p[pi])
    return calcNext(si + 1, pi + 1);
  else
    return calcNext(si + 1, pi);
}
int dpt[2005][2005];
bool vis[2005][2005];
int dp(int i, int char_to_rem) {
  if (i == s.size()) {
    if (char_to_rem == 0)
      return 0;
    else
      return -99999999;
  }
  if (vis[i][char_to_rem]) return dpt[i][char_to_rem];
  vis[i][char_to_rem] = true;
  int x = dp(i + 1, char_to_rem), y = -99999999, z = -99999999;
  if (char_to_rem > 0) y = dp(i + 1, char_to_rem - 1);
  if ((nextm[i] != -1) && (nextm[i] - i - p.size()) <= char_to_rem)
    z = dp(nextm[i], char_to_rem - nextm[i] + i + p.size()) + 1;
  return dpt[i][char_to_rem] = max(max(x, y), z);
}
int main() {
  cin >> s >> p;
  for (int i = 0; i < s.size(); ++i) nextm[i] = calcNext(i, 0);
  for (int i = 0; i < s.size(); ++i) cout << dp(0, i) << " ";
  cout << 0;
}
