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 N = 2007;
char st[N], p[N];
int n, np;
int dp[N][N];
void solve() {
  scanf("%s", st);
  n = strlen(st);
  scanf("%s", p);
  np = strlen(p);
  for (int i = 0; i < n; ++i) {
    int pos = i;
    int cnt = 0;
    int x = 0;
    while (cnt != np && pos >= 0) {
      if (st[pos] == p[np - cnt - 1])
        ++cnt;
      else
        ++x;
      --pos;
    }
    for (int j = 0; j <= i + 1; ++j) {
      if (i) dp[i][j] = dp[i - 1][j];
      if (cnt == np && j >= x) {
        if (pos == -1) {
          if (j == x) dp[i][j] = max(dp[i][j], 1);
        } else {
          if (j - x <= pos + 1) dp[i][j] = max(dp[i][j], dp[pos][j - x] + 1);
        }
      }
    }
  }
  for (int i = 0; i <= n; ++i) printf("%d ", dp[n - 1][i]);
  puts("");
}
int main() {
  cerr << fixed;
  cerr.precision(5);
  solve();
  return 0;
}
