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 maxn = 2010;
int dp[maxn][maxn];
int cnt[maxn];
int nxt[maxn];
namespace {
void run() {
  ios_base::sync_with_stdio(0);
  string s, p;
  cin >> s >> p;
  int n = s.size(), m = p.size();
  int i, j, t;
  fill(cnt, cnt + maxn, -1);
  for (i = 0; i < n; ++i)
    if (s[i] == p[0]) {
      t = 1;
      for (j = i + 1; j < n && t < m; ++j)
        if (s[j] == p[t]) ++t;
      if (t == m) {
        cnt[i] = j - i - m;
        nxt[i] = j;
      }
    }
  const int minf = -1000000000;
  for (i = n - 1; i >= 0; --i)
    for (j = 0; j <= n; ++j) {
      if (n - i < j) {
        dp[i][j] = minf;
        continue;
      }
      dp[i][j] = 0;
      if (n - 1 - i >= j && i < n - 1) dp[i][j] = max(0, dp[i + 1][j]);
      if (j > 0 && n - 1 - i >= j - 1 && i < n - 1)
        dp[i][j] = max(dp[i][j], dp[i + 1][j - 1]);
      if (cnt[i] != -1 && cnt[i] <= j) {
        if (n - nxt[i] >= j - cnt[i])
          dp[i][j] = max(dp[i][j], 1 + dp[nxt[i]][j - cnt[i]]);
      }
    }
  for (i = 0; i <= n; ++i) cout << dp[0][i] << ' ';
  cout << endl;
}
}  // namespace
int main() {
  run();
  return 0;
}
