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;
const int INF = 1000000000;
const int N = 2020;
int dp[N][N];
int main() {
  string s, p;
  cin >> s >> p;
  int n = s.size(), m = p.size();
  int max_rep = n / m;
  for (int i = 0; i < N; ++i)
    for (int j = 0; j < N; ++j) dp[i][j] = INF;
  dp[0][0] = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < N; ++j) dp[i + 1][j] = dp[i][j];
    int j = i;
    int pos = m - 1;
    while (j >= 0 && pos >= 0) {
      if (s[j] == p[pos]) --pos;
      --j;
    }
    if (pos == -1) {
      ++j;
      for (int k = max_rep - 1; k >= 0; --k) {
        dp[i + 1][k + 1] = min(dp[i + 1][k + 1], dp[j][k] + i - j + 1 - m);
      }
    }
  }
  int mx = 0;
  for (; mx <= max_rep; ++mx)
    if (dp[n][mx] == INF) break;
  --mx;
  int pp = 0;
  for (int c = 0; c <= dp[n][mx]; ++c) {
    while (dp[n][pp] <= c) ++pp;
    cout << pp - 1 << endl;
  }
  for (int c = dp[n][mx] + 1; c <= n; ++c) {
    cout << min(mx, (n - c) / m) << endl;
  }
  return 0;
}
