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 nxt[2005], cnt[2005], dp[2005][2005], n, m;
char a[2005], b[2005];
inline void Make(int p) {
  if (a[p] != b[1]) return;
  int i, j;
  for (i = p + 1, j = 2; i <= n && j <= m; ++i)
    if (a[i] != b[j])
      ++cnt[p];
    else
      ++j;
  if (j <= m) return;
  nxt[p] = i;
}
int main() {
  int i, j;
  cin.sync_with_stdio(false);
  cin >> (a + 1) >> (b + 1);
  n = strlen(a + 1);
  m = strlen(b + 1);
  for (i = 1; i <= n; ++i) Make(i);
  for (i = 0; i <= n + 1; ++i)
    for (j = 0; j <= n + 1; ++j) dp[i][j] = -1000000000;
  dp[n + 1][0] = 0;
  for (i = n; i; --i)
    for (j = 0; j <= n - i + 1; ++j) {
      dp[i][j] = dp[i + 1][j];
      if (j) dp[i][j] = max(dp[i][j], dp[i + 1][j - 1]);
      if (nxt[i] && cnt[i] <= j)
        dp[i][j] = max(dp[i][j], dp[nxt[i]][j - cnt[i]] + 1);
    }
  for (i = 0; i <= n; ++i) cout << dp[1][i] << " ";
  cout << "\n";
  return 0;
}
