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 <class T>
inline T bigmod(T p, T e, T M) {
  long long ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p) % M;
    p = (p * p) % M;
  }
  return (T)ret;
}
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T modinverse(T a, T M) {
  return bigmod(a, M - 2, M);
}
string S, P;
int dp[2001][2002], dp1[2001][2011], len[2001];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long a, b, c, d = 0, e, f = 0, g, h = 0, x, y, z = 0;
  cin >> S >> P;
  a = S.size(), b = P.size();
  for (int i = (0); i < (b); ++i) dp1[a][i] = 1000000;
  for (int i = a - 1; i >= 0; i--) {
    for (int j = 0; j < b; j++) {
      if (S[i] == P[j]) {
        dp1[i][j] = dp1[i + 1][j + 1];
      } else
        dp1[i][j] = 1 + dp1[i + 1][j];
    }
    len[i] = dp1[i][0];
  }
  for (int i = 1; i <= a; i++) dp[a][i] = -100000;
  for (int i = a - 1; i >= 0; i--) {
    for (int j = 0; j <= a; j++) {
      dp[i][j] = -100000;
      if (j > (a - i)) {
        continue;
      }
      dp[i][j] = max(dp[i][j], dp[i + 1][j]);
      if (j > 0) dp[i][j] = max(dp[i][j], dp[i + 1][j - 1]);
      if (len[i] <= j && (i + len[i] + b) <= a)
        dp[i][j] = max(dp[i][j], 1 + dp[i + len[i] + b][j - len[i]]);
    }
  }
  for (int i = 0; i <= a; i++) cout << dp[0][i] << " ";
  cout << endl;
}
