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 long long int MOD = 1e9 + 7;
long long int MulMod(long long int a, long long int b) { return (a * b) % MOD; }
long long int AddMod(long long int a, long long int b) { return (a + b) % MOD; }
const long long int MAXN = 2e3 + 10;
const long long int INF = MOD;
string second, p;
long long int n, m;
long long int pos[MAXN], dp[MAXN][MAXN];
long long int rec(long long int idx, long long int subLeft) {
  if (subLeft < 0) return -INF;
  if (subLeft > idx) return -INF;
  if (idx == 0) return 0;
  long long int &ans = dp[idx][subLeft];
  if (ans != -1) return ans;
  ans = max(rec(idx - 1, subLeft - 1), rec(idx - 1, subLeft));
  if (pos[idx - 1] != -1) {
    long long int len = idx - pos[idx - 1];
    ans = max(ans, 1 + rec(pos[idx - 1], subLeft - (len - m)));
  }
  return ans;
}
void solving() {
  cin >> second >> p;
  n = (long long int)second.size(), m = (long long int)p.size();
  memset((pos), (long long int)(-1), sizeof((pos)));
  memset((dp), (long long int)(-1), sizeof((dp)));
  for (long long int i = (0); i <= (n - 1); i++) {
    if (second[i] != p[m - 1]) continue;
    long long int j = i, k = m - 1;
    while (j >= 0 && k >= 0) {
      if (second[j] == p[k]) k--;
      j--;
    }
    if (k == -1) pos[i] = j + 1;
  }
  for (long long int i = (0); i <= (n); i++) cout << rec(n, i) << " ";
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int t = 1;
  for (long long int i = (1); i <= (t); i++) {
    solving();
  }
}
