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 double Pi = acos(-1.0);
string a, b;
int dp[2011][2011], len[2011];
int main() {
  if (fopen("data.in", "r")) freopen("data.in", "r", stdin);
  cin >> a;
  cin >> b;
  int l1 = a.size();
  int l2 = b.size();
  for (int i = 0; i < l1; i++) {
    int pos = i, cur = 0;
    while (pos < l1) {
      if (a[pos] == b[cur]) cur++;
      if (cur == l2) break;
      pos++;
    }
    if (cur == l2)
      len[i] = pos - i + 1;
    else
      len[i] = -1;
  }
  for (int i = 0; i < l1; i++) {
    for (int x = 0; x <= i; x++) {
      dp[i + 1][x + 1] = max(dp[i + 1][x + 1], dp[i][x]);
      dp[i + 1][x] = max(dp[i + 1][x], dp[i][x]);
      if (len[i] != -1)
        dp[i + len[i]][x + len[i] - l2] =
            max(dp[i + len[i]][x + len[i] - l2], dp[i][x] + 1);
    }
  }
  for (int i = 0; i <= l1; i++) cout << dp[l1][i] << " ";
  cout << endl;
  return 0;
}
