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 = 2005;
int dp[MAXn][MAXn];
const int INF = 1e9;
int ans[MAXn];
int main() {
  ios::sync_with_stdio(false);
  string s, p;
  cin >> s >> p;
  int n = s.size();
  int m = p.size();
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= n; j++) dp[i][j] = INF;
  dp[0][0] = 0;
  for (int i = 0; i < n; i++)
    for (int j = 0; j <= n; j++) {
      if (dp[i][j] == INF) continue;
      int k = j % m;
      dp[i + 1][j] = min(dp[i + 1][j], dp[i][j] + 1);
      if (k == 0) dp[i + 1][j] = min(dp[i + 1][j], dp[i][j]);
      if (s[i] == p[k]) dp[i + 1][j + 1] = min(dp[i + 1][j + 1], dp[i][j]);
    }
  for (int i = 0; i <= n; i++) {
    int rem = dp[n][i];
    if (rem == INF) continue;
    int k = i / m;
    for (int j = rem; j <= n - m * k; j++) ans[j] = max(ans[j], k);
  }
  for (int i = 0; i <= n; i++) cout << ans[i] << " ";
  cout << '\n';
  return 0;
}
