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 int INF = 1e9;
const long long int INFLL = 1e18;
const double EPS = 1e-8;
const long long int MOD = 1000000007;
template <class T>
T &chmin(T &a, const T &b) {
  return a = min(a, b);
}
template <class T>
T &chmax(T &a, const T &b) {
  return a = max(a, b);
}
const int MAX_N = 2010;
string s, p;
int to[MAX_N];
int dp[MAX_N][MAX_N];
int ans[MAX_N];
int main() {
  cin >> s >> p;
  int n = s.length();
  int m = p.length();
  for (int i = (0); i < (int)(n); i++) to[i] = INF;
  for (int i = (0); i < (int)(n); i++) {
    if (s[i] != p[0]) continue;
    int c = 0;
    for (int j = (i); j < (int)(n); j++) {
      if (s[j] == p[c]) c++;
      if (c == m) {
        to[i] = j;
        break;
      }
    }
  }
  for (int i = (0); i < (int)(MAX_N); i++)
    for (int j = (0); j < (int)(MAX_N); j++) dp[i][j] = INF;
  dp[0][0] = 0;
  for (int i = (0); i < (int)(n); i++) {
    for (int j = (0); j < (int)(n); j++) {
      chmin(dp[i + 1][j], dp[i][j]);
      if (to[i] != INF)
        chmin(dp[to[i] + 1][j + 1], dp[i][j] + to[i] + 1 - i - m);
    }
  }
  for (int i = (0); i < (int)(n + 1); i++)
    if (dp[n][i] != INF) ans[dp[n][i]] = i;
  for (int i = (1); i < (int)(n + 1); i++) chmax(ans[i], ans[i - 1]);
  for (int i = (0); i < (int)(n + 1); i++) chmin(ans[i], (n - i) / m);
  for (int i = (0); i < (int)(n + 1); i++) {
    if (i != 0) printf(" ");
    printf("%d", ans[i]);
  }
  printf("\n");
  return 0;
}
