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;
template <class T>
void debug(T a, T b) {
  ;
}
template <class T>
void chmin(T& a, const T& b) {
  if (a > b) a = b;
}
template <class T>
void chmax(T& a, const T& b) {
  if (a < b) a = b;
}
namespace std {
template <class S, class T>
ostream& operator<<(ostream& out, const pair<S, T>& a) {
  out << '(' << a.first << ',' << a.second << ')';
  return out;
}
}  // namespace std
long long int readL() {
  long long int res;
  scanf("%I64d", &res);
  return res;
}
void printL(long long int res) { printf("%I64d", res); }
const int INF = 5e8;
char s[2005], p[505];
int n, m;
int nxtpos[2005];
int dp[2005][2005];
int main() {
  scanf("%s%s", s, p);
  n = strlen(s);
  m = strlen(p);
  for (int i = 0; i < (n); ++i) {
    int k = 0;
    nxtpos[i] = INF;
    for (int j = (i); j < (n); ++j) {
      if (s[j] == p[k]) ++k;
      if (k == m) {
        nxtpos[i] = j + 1;
        break;
      }
    }
  }
  memset(dp, -1, sizeof(dp));
  dp[0][0] = 0;
  for (int i = 0; i < (n); ++i)
    for (int j = 0; j < (n + 1); ++j)
      if (dp[i][j] >= 0) {
        int nxt = nxtpos[i];
        int ers = nxt - i - m;
        if (nxt < INF) {
          chmax(dp[nxt][j + ers], dp[i][j] + 1);
        }
        chmax(dp[i + 1][j], dp[i][j]);
        chmax(dp[i + 1][j + 1], dp[i][j]);
      }
  for (int i = 0; i < (n + 1); ++i) {
    printf("%d%c", dp[n][i], i == n ? '\n' : ' ');
  }
  return 0;
}
