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;
int next() {
  int x;
  cin >> x;
  return x;
}
long long nextl() {
  long long x;
  cin >> x;
  return x;
}
double nextd() {
  double x;
  cin >> x;
  return x;
}
void out(long long x) { printf("%d ", x); }
void outln(long long x) { printf("%d\n", x); }
void outln() { printf("\n"); };
int main() {
  string s;
  string p;
  cin >> s;
  cin >> p;
  int n = s.length();
  int k = p.length();
  int next[n];
  for (int i = 0; i < n; i++) next[i] = -1;
  vector<pair<int, int> > pr;
  for (int i = 0; i < n; i++)
    if (s[i] == p[0]) {
      int start;
      int j;
      for (j = 0, start = i; j < k && start < n; start++)
        if (p[j] == s[start]) j++;
      if (j == k) pr.push_back((make_pair((i), (start))));
    }
  for (int i = 0; i < pr.size(); i++) next[pr[i].first] = pr[i].second;
  int mn[n + 1][n + 1];
  int inf = 1000000;
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= n; j++) mn[i][j] = inf;
  mn[0][0] = 0;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      mn[i + 1][j] = min(mn[i + 1][j], mn[i][j]);
      if (next[i] >= 0)
        mn[next[i]][j + 1] =
            min(mn[next[i]][j + 1], mn[i][j] + next[i] - i - k);
    }
  int best[n + 1];
  for (int i = 0; i <= n; i++) best[i] = 0;
  for (int i = 1; i <= n; i++)
    if (mn[n][i] < inf)
      for (int j = mn[n][i]; j <= n - i * k; j++) best[j] = i;
  for (int i = 0; i <= n; i++) out(best[i]);
  outln();
}
