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;
string s;
string p;
int a[2010];
int dp[2010][2010];
int main() {
  cin >> s;
  cin >> p;
  for (int i = 0; i < s.size(); i++) {
    int m = i;
    int j = 0;
    bool ok = 0;
    for (; j < p.size(); j++) {
      while (i < s.size() && s[i] != p[j]) {
        i++;
      }
      if (i == s.size() || s[i] != p[j]) {
        ok = 1;
        break;
      }
      i++;
    }
    if (ok == 1 || j < p.size()) {
      a[m] = -1;
      break;
    } else {
      a[m] = i - m;
    }
    i = m;
  }
  if (a[0] > 0) {
    dp[a[0] - 1][a[0] - p.size()] = 1;
  }
  for (int i = 0; i < s.size(); i++) {
    for (int j = 0; j <= i + 1; j++) {
      dp[i + 1][j] = max(dp[i][j], dp[i + 1][j]);
      dp[i + 1][j + 1] = max(dp[i + 1][j + 1], dp[i][j]);
      if (a[i + 1] > 0) {
        dp[i + a[i + 1]][j + a[i + 1] - p.size()] =
            max(dp[i][j] + 1, dp[i + a[i + 1]][j + a[i + 1] - p.size()]);
      }
    }
  }
  for (int i = 0; i < s.size(); i++) {
    if (i) {
      cout << " ";
    }
    cout << dp[s.size() - 1][i];
  }
  printf(" 0\n");
  return 0;
}
