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;
template <typename A, typename B>
inline bool mina(A &x, B y) {
  return (x > y) ? (x = y, 1) : 0;
}
template <typename A, typename B>
inline bool maxa(A &x, B y) {
  return (x < y) ? (x = y, 1) : 0;
}
char S[(2005)];
char P[(505)];
int E[(2005)];
int dp[(2005)][(2005)];
int main() {
  cin >> S >> P;
  int s = strlen(S);
  int p = strlen(P);
  for (int _b = (s), i = (0); i < _b; ++i) {
    int j = i;
    int k = 0;
    while (j < s && k < p) {
      if (S[j] == P[k]) ++k;
      ++j;
    }
    if (k == p) {
      E[i] = j;
    } else
      E[i] = -1;
  }
  for (int _b = (s + 1), i = (0); i < _b; ++i)
    for (int _b = (s + 1), j = (0); j < _b; ++j) dp[i][j] = -1e6;
  dp[0][0] = 0;
  for (int _b = (s), i = (0); i < _b; ++i)
    for (int _b = (s + 1), j = (0); j < _b; ++j)
      if (dp[i][j] >= 0) {
        if (E[i] == -1)
          maxa(dp[s][j], dp[i][j]);
        else if (j + E[i] - i - p <= s)
          maxa(dp[E[i]][j + E[i] - i - p], dp[i][j] + 1);
        maxa(dp[i + 1][j], dp[i][j]);
        maxa(dp[i + 1][j + 1], dp[i][j]);
      }
  for (int _b = (s + 1), i = (0); i < _b; ++i) printf("%d ", dp[s][i]);
  cout << endl;
  return 0;
}
