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 dp[2010][2010] = {};
char a[2010] = {};
char p[2010] = {};
int n, m;
int ff(int x) {
  if (x < m) return 2147483600;
  int t = m;
  int b = x;
  int tp = 0;
  while (t && b) {
    if (p[t] == a[b])
      t--, b--;
    else
      tp++, b--;
  }
  if (t == 0) {
    return x - b - m;
  } else {
    return 2147483600;
  }
}
int main() {
  scanf("%s", a + 1);
  scanf("%s", p + 1);
  n = strlen(a + 1);
  m = strlen(p + 1);
  for (int i = 0; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) dp[i][j] = -2147483600;
  }
  for (int i = 1; i <= n; i++) {
    int k = ff(i);
    for (int j = 0; j <= n; j++) {
      dp[i][j] = max(dp[i][j], dp[i - 1][j]);
    }
    for (int j = 0; j <= n; j++) {
      if (j >= k) dp[i][j] = max(dp[i][j], dp[i - k - m][j - k] + 1);
    }
  }
  for (int i = 0; i <= n; i++) {
    printf("%d", dp[n][i]);
    if (i != n) printf(" ");
  }
  return 0;
}
