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 sr[2014], p[2014];
int A[2014];
int dp[2014][2014];
inline void checkmax(int &x, int val) {
  if (x < val) x = val;
}
int main() {
  scanf("%s", sr);
  scanf("%s", p);
  int N = strlen(sr), L = strlen(p);
  for (int i = 0, j, k; i < N; i++) {
    for (j = i, k = 0; j < N; j++) {
      if (sr[j] == p[k]) k++;
      if (k == L) break;
    }
    if (j < N)
      A[i] = j;
    else
      A[i] = -1;
  }
  if (~A[0]) dp[A[0]][A[0] + 1 - L] = 1;
  for (int i = 0; i < N - 1; i++) {
    for (int j = 0; j <= i; j++) {
      if (~A[i + 1]) checkmax(dp[A[i + 1]][j + A[i + 1] - i - L], dp[i][j] + 1);
      checkmax(dp[i + 1][j], dp[i][j]);
      checkmax(dp[i + 1][j + 1], dp[i][j]);
    }
  }
  for (int i = 0; i <= N; i++) printf("%d%c", dp[N - 1][i], " \n"[i == N]);
  return 0;
}
