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;
int const MAX_N = 2010;
int const INF = 1e9;
int n, m, dp[MAX_N][MAX_N], cnt[MAX_N], answer[MAX_N];
string s, p;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> s >> p;
  n = (int)s.size();
  m = (int)p.size();
  fill(cnt, cnt + MAX_N, -1);
  for (int i = 0; i < n; ++i) {
    int j = i, k = 0;
    while (j < n) {
      if (s[j] != p[k]) {
        ++j;
      } else {
        ++j;
        ++k;
      }
      if (k == m) {
        cnt[i] = j - i;
        break;
      }
    }
  }
  for (int i = n - 1; i >= 0; --i) {
    if (cnt[i] == -1) {
      for (int j = 0; j < MAX_N; ++j) dp[i][j] = INF;
    } else {
      if (i + 1 < n)
        for (int j = 0; j < n; ++j) dp[i][j] = dp[i + 1][j];
      dp[i][1] = min(dp[i][1], cnt[i] - m);
      if (i + cnt[i] < n)
        for (int j = 1; j + 1 < n; ++j)
          dp[i][j + 1] = min(dp[i][j + 1], cnt[i] - m + dp[i + cnt[i]][j]);
    }
  }
  answer[0] = 0;
  for (int i = 0, j = 0; i < n; ++i) {
    if (s[i] == p[j]) {
      ++j;
    } else {
      j = 0;
    }
    if (j == m) {
      ++answer[0];
      j = 0;
    }
  }
  for (int i = 0; i < n; ++i)
    if (dp[0][i] != INF) answer[dp[0][i]] = max(answer[dp[0][i]], i);
  for (int i = 1; i < n; ++i)
    if (answer[i] == INF || answer[i - 1] > answer[i])
      answer[i] = answer[i - 1];
  for (int i = n - 1; i >= 0; --i) {
    if (i == 0 || answer[i - 1] < answer[i]) {
      int f = answer[i] * m;
      int e = n - i - f;
      for (int j = i; j < n; ++j) {
        answer[j] = min(answer[j], f / m);
        if (e > 0) {
          --e;
        } else {
          --f;
        }
      }
      break;
    }
  }
  for (int i = 0; i <= n; ++i) cout << answer[i] << ' ';
  cout << endl;
  return 0;
}
