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 rei() {
  int x;
  cin >> x;
  return x;
}
inline long long rel() {
  long long x;
  cin >> x;
  return x;
}
inline string res() {
  string x;
  cin >> x;
  return x;
}
int DP[2001][2001];
int R[2000];
void Calc() {
  string S = res();
  string T = res();
  int N = S.length();
  int M = T.length();
  for (int i = 0; i < N; i++) {
    int p = 0;
    int j = i;
    while (p < M && j < N) {
      if (S[j] == T[p]) {
        p++;
      }
      j++;
    }
    if (p == M) {
      R[i] = j;
    } else {
      R[i] = N + 1;
    }
  }
  for (int i = 0; i <= N; i++) {
    for (int j = 0; j <= N; j++) {
      DP[i][j] = -100000000;
    }
  }
  DP[0][0] = 0;
  for (int i = 0; i < N; i++) {
    for (int j = 0; j <= N; j++) {
      DP[i + 1][j] = max(DP[i + 1][j], DP[i][j]);
      DP[i + 1][j + 1] = max(DP[i + 1][j + 1], DP[i][j]);
      if (R[i] <= N && j + ((R[i] - i) - M) <= N) {
        DP[R[i]][j + ((R[i] - i) - M)] =
            max(DP[R[i]][j + ((R[i] - i) - M)], DP[i][j] + 1);
      }
    }
  }
  for (int i = 0; i <= N; i++) {
    cout << DP[N][i];
    if (i == N) {
      cout << "\n";
    } else {
      cout << " ";
    }
  }
}
int main(int argc, char** argv) {
  ios::sync_with_stdio(false), cin.tie(0);
  cout.tie(0);
  Calc();
  return 0;
}
