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 N = 2000 + 100;
const int lg = 20;
const long long int mod = 1e9 + 7;
const long long int inf = 2e9 + 100;
int dp[N][N], pr[N];
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  string second, p;
  cin >> second >> p;
  int n = second.size(), m = p.size();
  for (int i = 0; i < n; i++) {
    pr[i + 1] = -1;
    int x = m - 1;
    for (int j = i; j >= 0; j--) {
      x -= (p[x] == second[j]);
      if (x == -1) {
        pr[i + 1] = j;
        break;
      }
    }
  }
  for (int j = 1; j <= n; j++) dp[0][j] = -inf;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= i; j++) {
      dp[i][j] = dp[i - 1][j];
      if (j > 0) dp[i][j] = max(dp[i][j], dp[i - 1][j - 1]);
      int x = i - pr[i] - m;
      if (pr[i] != -1 && j >= x) dp[i][j] = max(dp[i][j], 1 + dp[pr[i]][j - x]);
    }
    for (int j = i + 1; j <= n; j++) dp[i][j] = -inf;
  }
  for (int i = 0; i <= n; i++) cout << dp[n][i] << ' ';
  cout << endl;
  return 0;
}
