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;
char s[2005], p[505];
int n, m, a[2005], dp[2005][2005];
int f(int i, int x) {
  if (i == n) return x == 0 ? 0 : -9999999;
  if (dp[i][x] != 9999999) return dp[i][x];
  int ans = f(i + 1, x);
  if (x) ans = max(ans, f(i + 1, x - 1));
  if (a[i] != -1 && a[i] - m <= x)
    ans = max(ans, 1 + f(i + a[i], x - (a[i] - m)));
  return dp[i][x] = ans;
}
int main() {
  scanf("%s", s);
  n = strlen(s);
  scanf("%s", p);
  m = strlen(p);
  for (int i = 0; i < n; i++) {
    int j = i, k = 0;
    for (; j < n && k < m; j++)
      if (s[j] == p[k]) k++;
    if (k == m)
      a[i] = j - i;
    else
      a[i] = -1;
  }
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= n; j++) dp[i][j] = 9999999;
  for (int i = 0; i <= n; i++) printf("%d ", max(0, f(0, i)));
  printf("\n");
  return 0;
}
