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;
int f[2005][2005], a[2005];
string s, p;
int main() {
  cin >> s;
  cin >> p;
  int l = p.length();
  memset(f, 0, sizeof(f));
  for (int i = 0; i < s.length(); i++) {
    int now = 0, x;
    for (int j = i; j < s.length(); j++) {
      if (s[j] == p[now]) now++;
      if (now == l) {
        x = j;
        break;
      }
    }
    if (now == l)
      f[x][x - l + 1] = 1, a[i] = x;
    else
      a[i] = -1;
  }
  for (int i = 0; i < s.length(); i++)
    for (int j = 0; j <= i + 1; j++) {
      f[i + 1][j] = max(f[i + 1][j], f[i][j]);
      f[i + 1][j + 1] = max(f[i + 1][j + 1], f[i][j]);
      if (i != s.length() - 1 && a[i + 1] != -1)
        f[a[i + 1]][j + a[i + 1] - i - l] =
            max(f[a[i + 1]][j + a[i + 1] - i - l], f[i][j] + 1);
    }
  for (int i = 0; i <= s.length(); i++) printf("%d ", f[s.length() - 1][i]);
  printf("\n");
  return 0;
}
