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;
inline int abs(int a) {
  if (a > 0) return a;
  return -a;
}
inline int sq(int a) { return a * a; }
char s[2010];
char p[2010];
int n, m;
int dp[2010][2010];
int main() {
  scanf("%s%s", s, p);
  n = strlen(s);
  m = strlen(p);
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= n; j++) dp[i][j] = -1000000000;
  dp[0][0] = 0;
  int ind = -1;
  for (int i = 1; i <= n; i++) {
    int j = m - 1, k = i - 1;
    while (j >= 0 && k >= 0) {
      if (p[j] == s[k]) j--;
      k--;
    }
    int rem = i - 1 - k - m;
    k++;
    dp[i][0] = dp[i - 1][0];
    if (j < 0 && rem == 0) dp[i][0] = max(dp[i][0], dp[k][0] + 1);
    for (int l = 1; l <= n; l++) {
      dp[i][l] = dp[i - 1][l];
      if (j < 0 && rem <= l && rem >= 0)
        dp[i][l] = max(dp[i][l], dp[k][l - rem] + 1);
      dp[i][l] = max(dp[i][l], dp[i - 1][l - 1]);
    }
  }
  for (int i = 0; i <= n; i++) printf("%d ", dp[n][i]);
  printf("\n");
  return 0;
}
