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 s1, s2;
long long mn[2003], dp[2003][2003];
long long mnd(long long a) {
  int i = s2.size() - 1;
  while (i >= 0 && a >= 0) {
    if (s1[a] == s2[i]) {
      a--;
      i--;
    } else
      a--;
  }
  if (a == -1 && i > -1) return a - 1;
  return a + 1;
}
int main() {
  cin >> s1 >> s2;
  if (s2.size() > s1.size()) {
    for (int i = 0; i <= s1.size(); i++) cout << 0 << " ";
    return 0;
  }
  for (int i = 0; i < s1.size(); i++) mn[i + 1] = mnd(i) + 1;
  for (int i = 0; i <= s1.size() - s2.size(); i++) {
    long long j = 0, w = i;
    while (s1[w] == s2[j] && w < s1.size()) {
      w++;
      j++;
      if (j == s2.size()) {
        dp[0][w] = 1;
        j = 0;
        i = w - 1;
        break;
      }
    }
  }
  for (int i = 1; i <= s1.size(); i++) {
    if (dp[0][i])
      dp[0][i] = dp[0][i - 1] + 1;
    else
      dp[0][i] = dp[0][i - 1];
  }
  for (int i = 1; i <= s1.size(); i++) {
    for (int j = i + 1; j <= s1.size(); j++) {
      if (mn[j] == -1)
        dp[i][j] = dp[i][j - 1];
      else if (j - mn[j] + 1 - s2.size() <= i && mn[j] - 1 >= 0 &&
               mn[j] - 1 >= i - (j - mn[j] + 1 - s2.size())) {
        dp[i][j] = dp[i - (j - mn[j] + 1 - s2.size())][mn[j] - 1] + 1;
        dp[i][j] = max(dp[i][j], dp[i][j - 1]);
      } else
        dp[i][j] = dp[i][j - 1];
      ;
    }
  }
  for (int i = 0; i <= s1.size(); i++) {
    cout << dp[i][s1.size()] << " ";
  }
}
