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 maxn = 2e3 + 5;
int dp[maxn][maxn], l[maxn], ans[maxn];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  string s, p;
  cin >> s >> p;
  int n = s.size(), m = p.size();
  for (int i = 0; i < n; i++) {
    int j = i, k = m - 1;
    if (s[j] != p[k]) {
      l[i] = -1;
      continue;
    }
    while (j >= 0 && k >= 0) {
      if (s[j] == p[k]) {
        k--;
      }
      j--;
    }
    if (k <= -1) {
      l[i] = j + 1;
    } else {
      l[i] = -1;
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= n; j++) {
      if (j > i + 1) {
        dp[i][j] = -1e9;
        continue;
      }
      dp[i][j] = (i > 0 && j > 0) ? dp[i - 1][j - 1] : 0;
      dp[i][j] = (i > 0) ? max(dp[i][j], dp[i - 1][j]) : dp[i][j];
      if (l[i] >= 0) {
        int d = i - l[i] + 1 - m;
        if (j >= d && l[i] > 0) {
          dp[i][j] = max(dp[i][j], dp[l[i] - 1][j - d] + 1);
        } else if (l[i] == 0 && j == d) {
          dp[i][j] = 1;
        }
      }
      ans[j] = max(dp[i][j], ans[j]);
    }
  }
  for (int i = 0; i <= n; i++) {
    cout << ans[i] << " ";
  }
  return 0;
}
