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 = 2005;
char s[N], t[N];
int dp[N][N], n, m;
int calc(int pos) {
  if (pos < m) return -1;
  int ans = 0, i = pos, j = m;
  while (i && j)
    if (s[i] == t[j])
      --i, --j;
    else
      ++ans, --i;
  return j == 0 ? ans : -1;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> (s + 1) >> (t + 1);
  n = strlen(s + 1);
  m = strlen(t + 1);
  for (int i = 0; i <= n; i++)
    for (int j = i + 1; j <= n; j++) dp[i][j] = -10000;
  for (int i = 1; i <= n; i++) {
    int x = calc(i);
    for (int j = 0; j <= n; j++) {
      dp[i][j] = max(dp[i - 1][j], dp[i][j]);
      if (x != -1 && j >= x) dp[i][j] = max(dp[i][j], dp[i - m - x][j - x] + 1);
    }
  }
  for (int i = 0; i <= n; i++) cout << dp[n][i] << ' ';
  return 0;
}
