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 = (int)1e9;
int n, m;
string s, p;
int dp[2005][2005], nxt[2005], ans[2005];
int calc(int i, int j) {
  if (i == n) return (j == 0 ? 0 : inf);
  if (j == 0) return 0;
  int &res = dp[i][j];
  if (res != -1) return res;
  res = calc(i + 1, j);
  int ii = nxt[i];
  if (ii == inf) return res = inf;
  res = min(res, ii - i - m + calc(ii, j - 1));
  return res;
}
int main() {
  getline(cin, s);
  getline(cin, p);
  n = (int)s.size();
  m = (int)p.size();
  memset(dp, -1, sizeof dp);
  for (int i = 0; i < n; i++) {
    int l, r = 0;
    for (l = i; l < n && r < m; l++) {
      if (s[l] == p[r]) r++;
    }
    if (r != m)
      nxt[i] = inf;
    else
      nxt[i] = l;
  }
  for (int i = 0; i <= n; i++) {
    int val = calc(0, i);
    if (val == inf) break;
    for (int j = val; j <= n - i * m; j++) ans[j] = i;
  }
  for (int i = 0; i <= n; i++) cout << ans[i] << " ";
  return 0;
}
