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;
using ll = long long;
using pii = pair<int, int>;
using tpl = tuple<int, int, int>;
using pcc = pair<char, char>;
const ll inf = 1e18 + 100;
const ll mod = 1e9 + 7;
int dr[] = {-1, 0, 1, 0};
int dc[] = {0, 1, 0, -1};
int kr[] = {-1, 1, -2, -2, -1, 1, 2, 2};
int kc[] = {-2, -2, -1, 1, 2, 2, -1, 1};
int dgr[] = {-1, -1, 1, 1};
int dgc[] = {1, -1, -1, 1};
const int N = 2e3 + 10;
int dp[N][N], ace[N];
int n, m;
string txt, ptn;
int fun(int idx, int cnt) {
  if (idx == n) {
    if (cnt == 0) return 0;
    return -2222;
  }
  if (cnt < 0) return -2222;
  int &mami = dp[idx][cnt];
  if (mami != -1) return mami;
  int ans = max(fun(idx + 1, cnt), fun(idx + 1, cnt - 1));
  if (ace[idx]) {
    int nxt = idx + ace[idx];
    ans = max(ans, 1 + fun(nxt, cnt - (ace[idx] - m)));
  }
  return mami = ans;
}
void solve() {
  cin >> txt >> ptn;
  n = txt.size();
  m = ptn.size();
  for (int i = 0; i < n; i++) {
    int idx = 0;
    for (int j = i; j < n; j++) {
      if (txt[j] == ptn[idx]) idx++;
      if (idx == m) {
        ace[i] = j - i + 1;
        break;
      }
    }
  }
  memset(dp, -1, sizeof dp);
  for (int i = 0; i <= n; i++) printf("%d ", fun(0, i));
  printf("\n");
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int tt = 1;
  while (tt--) {
    solve();
  }
  return 0;
}
